Directly related questions
-
20N.2.sl.TZ0.2d:
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
20N.2.sl.TZ0.2d:
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
20N.2.sl.TZ0.d:
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
20N.2.hl.TZ0.2f(i):
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
20N.2.hl.TZ0.f(i):
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
20N.2.hl.TZ0.2f(i):
The equilibrium constant, , for the conversion of A to B is in water at .
Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.
-
17N.2.hl.TZ0.6a.i:
State the equilibrium constant expression, Kc , for this reaction.
-
17N.2.hl.TZ0.6a.i:
State the equilibrium constant expression, Kc , for this reaction.
-
17N.2.hl.TZ0.a.i:
State the equilibrium constant expression, Kc , for this reaction.
- 17N.1.hl.TZ0.19: The enthalpy change for the dissolution of NH4NO3 is +26 kJ mol–1 at 25 °C. Which statement...
- 17N.1.hl.TZ0.19: The enthalpy change for the dissolution of NH4NO3 is +26 kJ mol–1 at 25 °C. Which statement...
-
21M.1.sl.TZ1.18:
Which changes produce the greatest increase in the percentage conversion of methane?
CH4 (g) + H2O (g) CO (g) + 3H2 (g)
-
21M.1.sl.TZ1.18:
Which changes produce the greatest increase in the percentage conversion of methane?
CH4 (g) + H2O (g) CO (g) + 3H2 (g)
- 21M.1.sl.TZ2.18: What effect does a catalyst have on the position of equilibrium and the value of the equilibrium...
- 21M.1.sl.TZ2.18: What effect does a catalyst have on the position of equilibrium and the value of the equilibrium...
-
21M.2.hl.TZ1.4d(iii):
State the expression for Kc for this stage of the reaction.
-
21M.2.hl.TZ1.4d(iii):
State the expression for Kc for this stage of the reaction.
-
21M.2.hl.TZ1.d(iii):
State the expression for Kc for this stage of the reaction.
- 21M.2.hl.TZ1.4d(iv): State and explain the effect of increasing temperature on the value of Kc.
- 21M.2.hl.TZ1.4d(iv): State and explain the effect of increasing temperature on the value of Kc.
- 21M.2.hl.TZ1.d(iv): State and explain the effect of increasing temperature on the value of Kc.
-
21M.2.sl.TZ2.2e:
Consider the following equilibrium reaction:
2SO2 (g) + O2 (g) 2SO3 (g)
State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.
-
21M.2.sl.TZ2.2e:
Consider the following equilibrium reaction:
2SO2 (g) + O2 (g) 2SO3 (g)
State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.
-
21M.2.sl.TZ2.e:
Consider the following equilibrium reaction:
2SO2 (g) + O2 (g) 2SO3 (g)
State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.
-
21M.2.hl.TZ2.7a:
State the equilibrium constant expression, Kc, for the reaction above.
-
21M.2.hl.TZ2.7a:
State the equilibrium constant expression, Kc, for the reaction above.
-
21M.2.hl.TZ2.a:
State the equilibrium constant expression, Kc, for the reaction above.
-
18M.1.hl.TZ2.23:
What occurs when the pressure on the given equilibrium is increased at constant temperature?
N2(g) + O2(g) 2NO(g) ΔH = +180 kJ
A. Kc increases and the position of equilibrium moves to the right.
B. Kc stays the same and the position of equilibrium is unchanged.
C. Kc stays the same and the position of equilibrium moves to the left.
D. Kc decreases and the position of equilibrium moves to the left.
-
18M.1.hl.TZ2.23:
What occurs when the pressure on the given equilibrium is increased at constant temperature?
N2(g) + O2(g) 2NO(g) ΔH = +180 kJ
A. Kc increases and the position of equilibrium moves to the right.
B. Kc stays the same and the position of equilibrium is unchanged.
C. Kc stays the same and the position of equilibrium moves to the left.
D. Kc decreases and the position of equilibrium moves to the left.
-
18M.1.hl.TZ1.22:
What is the effect of increasing the temperature in this reaction?
CO2(g) + H2O(l) H+(aq) + HCO3−(aq) ΔH < 0
A. The pH will decrease.
B. The pH will increase.
C. CO2 pressure will decrease.
D. The equilibrium position will shift to the right.
-
18M.1.hl.TZ1.22:
What is the effect of increasing the temperature in this reaction?
CO2(g) + H2O(l) H+(aq) + HCO3−(aq) ΔH < 0
A. The pH will decrease.
B. The pH will increase.
C. CO2 pressure will decrease.
D. The equilibrium position will shift to the right.
-
18M.2.hl.TZ1.1d.i:
State the equilibrium constant expression, Kc.
-
18M.2.hl.TZ1.1d.i:
State the equilibrium constant expression, Kc.
-
18M.2.hl.TZ1.d.i:
State the equilibrium constant expression, Kc.
-
18M.2.hl.TZ1.1d.ii:
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.2.hl.TZ1.1d.ii:
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.2.hl.TZ1.d.ii:
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.2.hl.TZ2.6c.i:
Nitrogen oxide is in equilibrium with dinitrogen dioxide.
2NO(g) N2O2(g) ΔHΘ < 0
Deduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.
-
18M.2.hl.TZ2.6c.i:
Nitrogen oxide is in equilibrium with dinitrogen dioxide.
2NO(g) N2O2(g) ΔHΘ < 0
Deduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.
-
18M.2.hl.TZ2.c.i:
Nitrogen oxide is in equilibrium with dinitrogen dioxide.
2NO(g) N2O2(g) ΔHΘ < 0
Deduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.
-
18M.1.sl.TZ1.18:
The equilibrium constant for N2(g) + 3H2(g) 2NH3(g) is K.
What is the equilibrium constant for this equation?
2N2(g) + 6H2(g) 4NH3(g)
A. K
B. 2K
C. K2
D. 2K2
-
18M.1.sl.TZ1.18:
The equilibrium constant for N2(g) + 3H2(g) 2NH3(g) is K.
What is the equilibrium constant for this equation?
2N2(g) + 6H2(g) 4NH3(g)
A. K
B. 2K
C. K2
D. 2K2
-
18M.2.sl.TZ1.1d:
Urea can also be made by the direct combination of ammonia and carbon dioxide gases.
2NH3(g) + CO2(g) (H2N)2CO(g) + H2O(g) ΔH < 0
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.2.sl.TZ1.1d:
Urea can also be made by the direct combination of ammonia and carbon dioxide gases.
2NH3(g) + CO2(g) (H2N)2CO(g) + H2O(g) ΔH < 0
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.2.sl.TZ1.d:
Urea can also be made by the direct combination of ammonia and carbon dioxide gases.
2NH3(g) + CO2(g) (H2N)2CO(g) + H2O(g) ΔH < 0
Predict, with a reason, the effect on the equilibrium constant, Kc, when the temperature is increased.
-
18M.1.sl.TZ2.18:
Which factor does not affect the position of equilibrium in this reaction?
2NO2(g) N2O4(g) ΔH = −58 kJ mol−1
A. Change in volume of the container
B. Change in temperature
C. Addition of a catalyst
D. Change in pressure
-
18M.1.sl.TZ2.18:
Which factor does not affect the position of equilibrium in this reaction?
2NO2(g) N2O4(g) ΔH = −58 kJ mol−1
A. Change in volume of the container
B. Change in temperature
C. Addition of a catalyst
D. Change in pressure
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18M.2.sl.TZ2.5a:
Distinguish between the terms reaction quotient, Q, and equilibrium constant, Kc.
-
18M.2.sl.TZ2.5a:
Distinguish between the terms reaction quotient, Q, and equilibrium constant, Kc.
-
18M.2.sl.TZ2.a:
Distinguish between the terms reaction quotient, Q, and equilibrium constant, Kc.
-
18M.2.sl.TZ2.5b:
The equilibrium constant, Kc, is 0.282 at temperature T.
Deduce, showing your work, the direction of the initial reaction.
-
18M.2.sl.TZ2.5b:
The equilibrium constant, Kc, is 0.282 at temperature T.
Deduce, showing your work, the direction of the initial reaction.
-
18M.2.sl.TZ2.b:
The equilibrium constant, Kc, is 0.282 at temperature T.
Deduce, showing your work, the direction of the initial reaction.
-
21N.1.sl.TZ0.19:
The equilibrium 2H2 (g) + N2 (g) N2H4 (g) has an equilibrium constant, K, at 150 °C.
What is the equilibrium constant at 150 °C, for the reverse reaction?
N2H4 (g) 2H2 (g) + N2 (g)
A. KB. K−1
C. −K
D. 2K
-
21N.1.sl.TZ0.19:
The equilibrium 2H2 (g) + N2 (g) N2H4 (g) has an equilibrium constant, K, at 150 °C.
What is the equilibrium constant at 150 °C, for the reverse reaction?
N2H4 (g) 2H2 (g) + N2 (g)
A. KB. K−1
C. −K
D. 2K
-
21N.1.hl.TZ0.22:
A reversible reaction has a reaction quotient, Q, of 4.5 and equilibrium constant, Kc, of 6.2.
2A (g) A2 (g)
Which statement describes the reaction at this time?
A. The system has reached equilibrium.B. The rate of the forward reaction is greater than the rate of the reverse reaction.
C. The concentration of reactant is greater than the concentration of product.
D. At equilibrium, the concentration of reactant is greater than the concentration of product.
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21N.1.hl.TZ0.22:
A reversible reaction has a reaction quotient, Q, of 4.5 and equilibrium constant, Kc, of 6.2.
2A (g) A2 (g)
Which statement describes the reaction at this time?
A. The system has reached equilibrium.B. The rate of the forward reaction is greater than the rate of the reverse reaction.
C. The concentration of reactant is greater than the concentration of product.
D. At equilibrium, the concentration of reactant is greater than the concentration of product.
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21N.2.sl.TZ0.3c(ii):
State the equilibrium constant expression, Kc, for this reaction.
-
21N.2.sl.TZ0.3c(ii):
State the equilibrium constant expression, Kc, for this reaction.
-
21N.2.sl.TZ0.c(ii):
State the equilibrium constant expression, Kc, for this reaction.
- 21N.2.sl.TZ0.3c(iii): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
- 21N.2.sl.TZ0.3c(iii): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
- 21N.2.sl.TZ0.c(iii): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
-
21N.2.hl.TZ0.3c(v):
State the equilibrium constant expression, Kc, for this reaction.
-
21N.2.hl.TZ0.3c(v):
State the equilibrium constant expression, Kc, for this reaction.
-
21N.2.hl.TZ0.c(v):
State the equilibrium constant expression, Kc, for this reaction.
- 21N.2.hl.TZ0.3c(vi): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
- 21N.2.hl.TZ0.3c(vi): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
- 21N.2.hl.TZ0.c(vi): State, with a reason, the effect of an increase in temperature on the position of this equilibrium.
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18N.2.hl.TZ0.5e:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
18N.2.hl.TZ0.5e:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
18N.2.hl.TZ0.e:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
18N.1.sl.TZ0.18:
Consider the reaction:
2N2O (g) 2N2 (g) + O2 (g)
The values of Kc at different temperatures are:
Which statement is correct at higher temperature?
A. The forward reaction is favoured.
B. The reverse reaction is favoured.
C. The rate of the reverse reaction is greater than the rate of the forward reaction.
D. The concentration of both reactants and products increase.
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18N.1.sl.TZ0.18:
Consider the reaction:
2N2O (g) 2N2 (g) + O2 (g)
The values of Kc at different temperatures are:
Which statement is correct at higher temperature?
A. The forward reaction is favoured.
B. The reverse reaction is favoured.
C. The rate of the reverse reaction is greater than the rate of the forward reaction.
D. The concentration of both reactants and products increase.
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18N.2.sl.TZ0.5b:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
18N.2.sl.TZ0.5b:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
18N.2.sl.TZ0.b:
0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm3 flask at 1000 K.
Predict the direction of the reaction showing your working.
-
22M.1.hl.TZ1.23:
At equilibrium, the concentrations of chlorine and iodine are both 0.02 mol dm–3.
Cl2 (g) + (g) Cl (g) Kc = 454
What is the concentration of iodine monochloride, Cl?
A.B.
C.
D.
-
22M.1.hl.TZ1.23:
At equilibrium, the concentrations of chlorine and iodine are both 0.02 mol dm–3.
Cl2 (g) + (g) Cl (g) Kc = 454
What is the concentration of iodine monochloride, Cl?
A.B.
C.
D.
-
22M.1.sl.TZ2.18:
The equilibrium constant, Kc, for the reaction 2A + 4B 2C + 4D has a value of 4.0. What is the value of Kc for the reaction below at the same temperature?
C + 2D A + 2B
A. 0.25
B. 0.50
C. 1.0
D. 16
-
22M.1.sl.TZ2.18:
The equilibrium constant, Kc, for the reaction 2A + 4B 2C + 4D has a value of 4.0. What is the value of Kc for the reaction below at the same temperature?
C + 2D A + 2B
A. 0.25
B. 0.50
C. 1.0
D. 16
-
22M.1.hl.TZ2.23:
0.50 mol of (g) and 0.50 mol of Br2 (g) are placed in a closed flask. The following equilibrium is established.
(g) + Br2 (g) 2Br (g)
The equilibrium mixture contains 0.80 mol of Br (g). What is the value of Kc?
A. 0.64
B. 1.3
C. 2.6
D. 64
-
22M.1.hl.TZ2.23:
0.50 mol of (g) and 0.50 mol of Br2 (g) are placed in a closed flask. The following equilibrium is established.
(g) + Br2 (g) 2Br (g)
The equilibrium mixture contains 0.80 mol of Br (g). What is the value of Kc?
A. 0.64
B. 1.3
C. 2.6
D. 64
-
22M.2.sl.TZ1.2c(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
-
22M.2.sl.TZ1.2c(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
-
22M.2.sl.TZ1.c(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
- 22M.2.sl.TZ1.2c(iii): State how the use of a catalyst affects the position of the equilibrium.
- 22M.2.sl.TZ1.2c(iii): State how the use of a catalyst affects the position of the equilibrium.
- 22M.2.sl.TZ1.c(iii): State how the use of a catalyst affects the position of the equilibrium.
-
22M.2.sl.TZ1.2c(ii):
Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, Kc.
-
22M.2.sl.TZ1.2c(ii):
Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, Kc.
-
22M.2.sl.TZ1.c(ii):
Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, Kc.
-
22M.2.hl.TZ1.3a(iii):
With reference to the reaction quotient, Q, explain why the percentage yield increases as the pressure is increased at constant temperature.
-
22M.2.hl.TZ1.a(iii):
With reference to the reaction quotient, Q, explain why the percentage yield increases as the pressure is increased at constant temperature.
-
22M.2.hl.TZ1.3a(iii):
With reference to the reaction quotient, Q, explain why the percentage yield increases as the pressure is increased at constant temperature.
-
22M.2.hl.TZ1.3b(iii):
Demonstrate that your answer to (b)(i) is consistent with the effect of an increase in temperature on the percentage yield, as shown in the graph.
-
22M.2.hl.TZ1.b(iii):
Demonstrate that your answer to (b)(i) is consistent with the effect of an increase in temperature on the percentage yield, as shown in the graph.
-
22M.2.hl.TZ1.3b(iii):
Demonstrate that your answer to (b)(i) is consistent with the effect of an increase in temperature on the percentage yield, as shown in the graph.
- 22M.2.hl.TZ1.3a(ii): State how the use of a catalyst affects the position of the equilibrium.
- 22M.2.hl.TZ1.a(ii): State how the use of a catalyst affects the position of the equilibrium.
- 22M.2.hl.TZ1.3a(ii): State how the use of a catalyst affects the position of the equilibrium.
- 22M.2.sl.TZ2.3b(ii): Explain the effect of increasing temperature on the yield of SO3.
- 22M.2.sl.TZ2.3b(ii): Explain the effect of increasing temperature on the yield of SO3.
- 22M.2.sl.TZ2.b(ii): Explain the effect of increasing temperature on the yield of SO3.
-
22M.2.hl.TZ2.4c:
State the equilibrium constant expression, Kc, for this reaction.
-
22M.2.hl.TZ2.4c:
State the equilibrium constant expression, Kc, for this reaction.
-
22M.2.hl.TZ2.c:
State the equilibrium constant expression, Kc, for this reaction.
-
19M.2.hl.TZ1.4c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) ⇌ CH3COOOH (aq) + H2O (l)
-
19M.2.hl.TZ1.4c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) ⇌ CH3COOOH (aq) + H2O (l)
-
19M.2.hl.TZ1.c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) ⇌ CH3COOOH (aq) + H2O (l)
-
19M.2.hl.TZ1.5b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.hl.TZ1.5b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.hl.TZ1.b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.hl.TZ2.5b:
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.2.hl.TZ2.5b:
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.2.hl.TZ2.b:
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.3.hl.TZ2.1c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19M.3.hl.TZ2.1c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19M.3.hl.TZ2.c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19M.3.hl.TZ2.1b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.3.hl.TZ2.1b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.3.hl.TZ2.b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.1.hl.TZ2.22:
Consider the following equilibrium reaction.
2N2O (g) + O2 (g) 4NO (g) ΔH = +16 kJ
Which change will move the equilibrium to the right?
A. Decrease in pressure
B. Decrease in temperature
C. Increase in [NO]
D. Decrease in [O2]
-
19M.1.hl.TZ2.22:
Consider the following equilibrium reaction.
2N2O (g) + O2 (g) 4NO (g) ΔH = +16 kJ
Which change will move the equilibrium to the right?
A. Decrease in pressure
B. Decrease in temperature
C. Increase in [NO]
D. Decrease in [O2]
-
19M.2.sl.TZ1.5b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.sl.TZ1.5b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.sl.TZ1.b:
A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.
Cl2 (g) + 2NaOH (aq) NaOCl (aq) + NaCl (aq) + H2O (l)
Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.
-
19M.2.sl.TZ2.5a(iii):
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.2.sl.TZ2.5a(iii):
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.2.sl.TZ2.a(iii):
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
-
19M.1.sl.TZ2.18:
What is the equilibrium constant expression for the following equation?
2NO2 (g) + F2 (g) 2NO2F (g)
A.
B.
C.
D.
-
19M.1.sl.TZ2.18:
What is the equilibrium constant expression for the following equation?
2NO2 (g) + F2 (g) 2NO2F (g)
A.
B.
C.
D.
-
19M.1.sl.TZ1.18:
Kc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
What is Kc for the following reaction, at the same temperature?
N2 (g) + O2 (g) N2O (g)
A. 7.3 × 1034
B.
C.
D.
-
19M.1.sl.TZ1.18:
Kc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
What is Kc for the following reaction, at the same temperature?
N2 (g) + O2 (g) N2O (g)
A. 7.3 × 1034
B.
C.
D.
-
19N.1.sl.TZ0.19:
What effect does increasing both pressure and temperature have on the equilibrium constant, Kc?
N2 (g) + 3H2 (g) 2NH3 (g) ΔH = −45.9 kJ
A. Decreases
B. Increases
C. Remains constant
D. Cannot be predicted as effects are opposite
-
19N.1.sl.TZ0.19:
What effect does increasing both pressure and temperature have on the equilibrium constant, Kc?
N2 (g) + 3H2 (g) 2NH3 (g) ΔH = −45.9 kJ
A. Decreases
B. Increases
C. Remains constant
D. Cannot be predicted as effects are opposite
- 19N.2.hl.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.hl.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.hl.TZ0.a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
-
22N.1.sl.TZ0.18:
The exothermic reaction 2 (g) + 3Cl2 (g) 2Cl3 (g) is at equilibrium in a fixed volume. What is correct about the reaction quotient, Q, and shift in position of equilibrium the instant temperature is raised?
A. Q > K, equilibrium shifts right towards products.B. Q > K, equilibrium shifts left towards reactants.
C. Q < K, equilibrium shifts right towards products.
D. Q < K, equilibrium shifts left towards reactants.
-
22N.1.sl.TZ0.18:
The exothermic reaction 2 (g) + 3Cl2 (g) 2Cl3 (g) is at equilibrium in a fixed volume. What is correct about the reaction quotient, Q, and shift in position of equilibrium the instant temperature is raised?
A. Q > K, equilibrium shifts right towards products.B. Q > K, equilibrium shifts left towards reactants.
C. Q < K, equilibrium shifts right towards products.
D. Q < K, equilibrium shifts left towards reactants.
-
22N.1.hl.TZ0.23:
Which of these changes would shift the equilibrium to the right?
[Co(H2O)6]2+ (aq) + 4Cl− (aq) [CoCl4]2− (aq) + 6H2O (l)
I. Addition of 0.01 M HCl
II. Addition of concentrated HCl
III. Evaporation of water
A. I and II onlyB. I and III only
C. II and III only
D. I, II and III
-
22N.1.hl.TZ0.23:
Which of these changes would shift the equilibrium to the right?
[Co(H2O)6]2+ (aq) + 4Cl− (aq) [CoCl4]2− (aq) + 6H2O (l)
I. Addition of 0.01 M HCl
II. Addition of concentrated HCl
III. Evaporation of water
A. I and II onlyB. I and III only
C. II and III only
D. I, II and III
- 22N.2.sl.TZ0.4a.ii: Explain, with reference to Le Châtelier’s principle, the effect of using dilute rather than...
- 22N.2.sl.TZ0.4a.ii: Explain, with reference to Le Châtelier’s principle, the effect of using dilute rather than...
- 22N.2.sl.TZ0.a.ii: Explain, with reference to Le Châtelier’s principle, the effect of using dilute rather than...
-
22N.2.hl.TZ0.1d.viii:
Deduce, with a reason, the position of the equilibrium.
-
22N.2.hl.TZ0.1d.viii:
Deduce, with a reason, the position of the equilibrium.
-
22N.2.hl.TZ0.d.viii:
Deduce, with a reason, the position of the equilibrium.
-
17N.1.sl.TZ0.18:
What will happen if the pressure is increased in the following reaction mixture at equilibrium?
CO2 (g) + H2O (l) H+ (aq) + HCO3− (aq)
A. The equilibrium will shift to the right and pH will decrease.
B. The equilibrium will shift to the right and pH will increase.
C. The equilibrium will shift to the left and pH will increase.
D. The equilibrium will shift to the left and pH will decrease.
-
17N.1.sl.TZ0.18:
What will happen if the pressure is increased in the following reaction mixture at equilibrium?
CO2 (g) + H2O (l) H+ (aq) + HCO3− (aq)
A. The equilibrium will shift to the right and pH will decrease.
B. The equilibrium will shift to the right and pH will increase.
C. The equilibrium will shift to the left and pH will increase.
D. The equilibrium will shift to the left and pH will decrease.
-
17N.2.sl.TZ0.5a:
The following reaction was allowed to reach equilibrium at 761 K.
H2 (g) + I2 (g) 2HI (g) ΔHθ < 0
Outline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.
-
17N.2.sl.TZ0.5a:
The following reaction was allowed to reach equilibrium at 761 K.
H2 (g) + I2 (g) 2HI (g) ΔHθ < 0
Outline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.
-
17N.2.sl.TZ0.a:
The following reaction was allowed to reach equilibrium at 761 K.
H2 (g) + I2 (g) 2HI (g) ΔHθ < 0
Outline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.
-
19M.2.hl.TZ2.5d(ii):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.2.hl.TZ2.5d(ii):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.2.hl.TZ2.d(ii):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.1.hl.TZ1.22:
Kc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
What is Kc for the following reaction, at the same temperature?
N2 (g) + O2 (g) N2O (g)
A. 7.3 × 1034
B.
C.
D.
-
19M.1.hl.TZ1.22:
Kc for 2N2O (g) 2N2 (g) + O2 (g) is 7.3 × 1034.
What is Kc for the following reaction, at the same temperature?
N2 (g) + O2 (g) N2O (g)
A. 7.3 × 1034
B.
C.
D.
-
19M.2.sl.TZ1.4c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) CH3COOOH (aq) + H2O (l)
-
19M.2.sl.TZ1.4c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) CH3COOOH (aq) + H2O (l)
-
19M.2.sl.TZ1.c:
Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.
H2O2 (aq) + CH3COOH (aq) CH3COOOH (aq) + H2O (l)
-
19M.2.sl.TZ2.5b(i):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.2.sl.TZ2.5b(i):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.2.sl.TZ2.b(i):
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
-
19M.3.sl.TZ2.1b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.3.sl.TZ2.1b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.3.sl.TZ2.b(i):
Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.
-
19M.3.sl.TZ2.1c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19M.3.sl.TZ2.1c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19M.3.sl.TZ2.c(i):
Lead(II) chloride, PbCl2, has very low solubility in water.
PbCl2 (s) Pb2+ (aq) + 2Cl− (aq)
Explain why the presence of chloride ions in beverages affects lead concentrations.
-
19N.2.hl.TZ0.4a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on Ka, of increasing the temperature.
-
19N.2.hl.TZ0.4a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on Ka, of increasing the temperature.
-
19N.2.hl.TZ0.a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on Ka, of increasing the temperature.
- 19N.2.sl.TZ0.4a(ii): The value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10−4. State,...
- 19N.2.sl.TZ0.4a(ii): The value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10−4. State,...
- 19N.2.sl.TZ0.a(ii): The value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10−4. State,...
-
19N.2.sl.TZ0.4a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on the equilibrium constant, of increasing the temperature.
-
19N.2.sl.TZ0.4a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on the equilibrium constant, of increasing the temperature.
-
19N.2.sl.TZ0.a(iii):
The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H+], and on the equilibrium constant, of increasing the temperature.
-
20N.1.sl.TZ0.18:
What is correct when temperature increases in this reaction at equilibrium?
-
20N.1.sl.TZ0.18:
What is correct when temperature increases in this reaction at equilibrium?
-
20N.2.hl.TZ0.5c:
State the expression for ethanoic acid.
-
20N.2.hl.TZ0.5c:
State the expression for ethanoic acid.
-
20N.2.hl.TZ0.c:
State the expression for ethanoic acid.
- 20N.3.sl.TZ0.2b: The ice bath is used at equilibrium to slow down the forward and reverse reactions. Explain why...
- 20N.3.sl.TZ0.2b: The ice bath is used at equilibrium to slow down the forward and reverse reactions. Explain why...
- 20N.3.sl.TZ0.b: The ice bath is used at equilibrium to slow down the forward and reverse reactions. Explain why...
- 20N.3.sl.TZ0.2c: Suggest why the titration must be conducted quickly even though a low temperature is maintained.
- 20N.3.sl.TZ0.2c: Suggest why the titration must be conducted quickly even though a low temperature is maintained.
- 20N.3.sl.TZ0.c: Suggest why the titration must be conducted quickly even though a low temperature is maintained.
-
21M.1.hl.TZ2.23:
Sulfur dioxide reacts with oxygen to form sulfur trioxide.
2SO2 (g) + O2 (g) 2SO3 (g) ΔH = −197 kJ
Which change increases the value of Kc?
A. increasing the temperature
B. decreasing the temperature
C. decreasing [SO2 (g)]
D. decreasing [SO3 (g)]
-
21M.1.hl.TZ2.23:
Sulfur dioxide reacts with oxygen to form sulfur trioxide.
2SO2 (g) + O2 (g) 2SO3 (g) ΔH = −197 kJ
Which change increases the value of Kc?
A. increasing the temperature
B. decreasing the temperature
C. decreasing [SO2 (g)]
D. decreasing [SO3 (g)]
-
21M.2.sl.TZ1.4d(ii):
State the expression for Kc for this stage of the reaction.
-
21M.2.sl.TZ1.4d(ii):
State the expression for Kc for this stage of the reaction.
-
21M.2.sl.TZ1.d(ii):
State the expression for Kc for this stage of the reaction.
- 21M.2.sl.TZ1.4d(iii): State and explain the effect of increasing temperature on the value of Kc.
- 21M.2.sl.TZ1.4d(iii): State and explain the effect of increasing temperature on the value of Kc.
- 21M.2.sl.TZ1.d(iii): State and explain the effect of increasing temperature on the value of Kc.
- 21M.2.hl.TZ2.7b: State and explain how the equilibrium would be affected by increasing the volume of the reaction...
- 21M.2.hl.TZ2.7b: State and explain how the equilibrium would be affected by increasing the volume of the reaction...
- 21M.2.hl.TZ2.b: State and explain how the equilibrium would be affected by increasing the volume of the reaction...
- 21N.1.hl.TZ0.27: What is correct for pure hot water?
- 21N.1.hl.TZ0.27: What is correct for pure hot water?
-
22M.1.sl.TZ1.18:
Cl2 (g) + (g) Cl (g) Kc = 454
What is the Kc value for the reaction below?
2 Cl (g) Cl2 (g) + (g)
A.
B.
C.
D.
-
22M.1.sl.TZ1.18:
Cl2 (g) + (g) Cl (g) Kc = 454
What is the Kc value for the reaction below?
2 Cl (g) Cl2 (g) + (g)
A.
B.
C.
D.
-
22M.1.sl.TZ1.19:
Which species are acids in the equilibrium below?
CH3NH2 + H2O CH3NH3+ + OH–
A. CH3NH2 and H2O
B. H2O and CH3NH3+
C. H2O and OH–
D. CH3NH2 and CH3NH3+
-
22M.1.sl.TZ1.19:
Which species are acids in the equilibrium below?
CH3NH2 + H2O CH3NH3+ + OH–
A. CH3NH2 and H2O
B. H2O and CH3NH3+
C. H2O and OH–
D. CH3NH2 and CH3NH3+
-
22M.2.hl.TZ1.3a(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
-
22M.2.hl.TZ1.3a(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
-
22M.2.hl.TZ1.a(i):
Deduce the expression for the equilibrium constant, Kc, for this equation.
- 22M.2.hl.TZ2.6b(ii): Explain the effect of increasing temperature on the yield of SO3.
- 22M.2.hl.TZ2.6b(ii): Explain the effect of increasing temperature on the yield of SO3.
- 22M.2.hl.TZ2.b(ii): Explain the effect of increasing temperature on the yield of SO3.