The best definition is A measure of the number of ways of arranging energy within particles and this is the correct answer.

Entropy is sometimes described as a 'measure of disorder', but this may be misleading and should be avoided.

'The change in heat energy at constant pressure' is a definition of Enthalpy change (ΔH). 'The average kinetic energy of a system' is a definition of temperature, T.

Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Any change that leads to a greater number of gaseous moles tends to lead to an increase in the entropy of the system.

Thus the correct answer here is CaCO_3(s) → CaO_(s) + CO_2(g)

The other reactions involve the creation of liquids or solids from gases, or a decrease in the number of gaseous moles.

Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Any change that leads to a greater number of gaseous moles tends to lead to an increase in the entropy of the system.

Thus the correct answer here is CH_4(g) + 1½O_2(g) → CO_(g) + 2H₂O_(g) since 2½ moles of gas react to give 3 moles of gas.

The other reactions involve the creation of liquid water rather than gaseous water, or no increase in the number of gaseous moles.

Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Any change that leads to a greater number of gaseous moles tends to lead to an increase in the entropy of the system.

Thus the correct answer is H_2(g) + Cl_2(g) → 2HCl_(g) since 2 moles of gas react to give 2 moles of gas. So the entropy change of the system is likely to be very close to zero.

The other reactions involve the creation of liquid water from gaseous water (negative ΔS), a decrease in the number of gaseous moles (negative ΔS), or the creation of a gaseous product from a solid reactant (positive ΔS).

The standard entropy change can be calculated by finding the difference between the standard entropy of the reactants and the standard entropy of the products; ΔS = S(products) – S(reactants).

The standard entropy of the reactants is 186 + 2×205.0 = 596.0

The standard entropy of the products is 213.8 + 2×70.0 = 353.8

Thus the standard entropy change is 353.8 – 596.0 = –242.2 (in JK^–1mol^–1) which is the correct answer.

Incorrect answers

+242.2 is obtained if ΔS is found by S(reactants) – S(products) instead of ΔS = S(products) – S(reactants).

–107.2 is obtained if one mole is used for every species in the calculation instead of using two moles of oxygen and two moles of water.

+107.2 is obtained if one mole is used for every species in the calculation instead of using two moles of oxygen and two moles of water and ΔS is found by S(reactants) – S(products) instead of ΔS = S(products) – S(reactants).

The entropy change can be calculated by finding the difference between the standard entropy of the reactants and the standard entropy of the products; ΔS = S(products) – S(reactants).

The standard entropy of the reactants is 2×131.0 + 205.0 = 467.0

The standard entropy of the products is 2×188.8 = 377.6

Thus the standard entropy change is 377.6 – 467.0 = –89.4 (in JK^–1mol^–1) which is the correct answer.

Incorrect answers

–278.2 is obtained if one mole of water is used (products) instead of two moles of water.

–147.2 is obtained if one mole is used for every species in the calculation instead of using two moles of hydrogen and two moles of water.

–327 is obtained if the value used for water is liquid water (+70.0) instead of gaseous water (+188.8) as written in the equation.

For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero (there must be an overall gain in entropy in the universe). Since ΔG = ΔH−TΔS (this should be learned), the correct answer is ΔH−TΔS ≤ 0 since this is stating that ΔG ≤ 0.

Incorrect answers (refer to ΔG = ΔH−TΔS)

The reaction does not always have to be exothermic. (ΔH < 0 shows an exothermic reaction.) Exothermic reactions will often be spontaneous, but ΔG may not be negative (or zero) if ΔS is negative. Endothermic reactions (positive ΔH) can be spontaneous; ΔG can be negative (or zero) if ΔS is positive.

If the entropy change of the system ΔS is positive this may lead to a spontaneous reaction; ΔG will be negative (or zero) if the reaction is also exothermic (negative ΔH). However, if the reaction is endothermic (positive ΔH) then ΔG may not be less than or equal to zero so the reaction will not be spontaneous.

For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero (there must be an overall gain in entropy in the universe). Since ΔG = ΔH−TΔS (this should be learned), the correct answer is the reaction will become spontaneous at higher temperature since at some point as temperature increases ΔG ≤ 0.

Looking carefully at the Gibb's free energy equation, ΔG = ΔH−TΔS, if ΔH is positive and ΔS is positive then ΔG will be negative when |TΔS| > |ΔH|; this will occur as temperature increases, as |TΔS| becomes greater as temperature becomes greater.

Note: |x| means magnitude of x, that is the numerical value of x without a sign.

Remember: Temperature, T in the expression is given in Kelvin (K) so is always positive.

Incorrect answers (refer to ΔG = ΔH−TΔS)

For the reaction to be become spontaneous at lower temperature, ΔH and ΔS would both need to be negative; then ΔG will be negative when |TΔS| < |ΔH|; this will occur as temperature decreases, as |TΔS| becomes smaller as temperature becomes smaller.

For the reaction to always be spontaneous (at any temperature), ΔH would need to be negative and ΔS would need to be positive; then ΔG will always be negative.

For the reaction to never be spontaneous (at any temperature), ΔH would need to be positive and ΔS would need to be negative; then ΔG will always be positive.

The standard free energy change (ΔG^o) can be calculated by using ΔG^o = ΔH^o–TΔS^o.

Temperature must be in K (as given in the question).

Units for enthalpy include kilojoules and units for entropy include joules. So it is best to convert entropy into kilojoules by dividing by 1000; −240 J K^–1 mol^–1 becomes −0.240 kJ K^–1 mol^–1.

Therefore ΔG^o = −287−(298×−0.240)

= −215.48 = –215 kJ mol^–1 to three sig figs

The standard free energy value is negative, so the reaction is spontaneous. Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero.

Thus the correct answer is –215; yes, it is spontaneous.

Incorrect answers

+71200 (+71233) is obtained if entropy value (−240) is not converted into kilojoules.

The standard free energy change (ΔG^o) can be calculated by using ΔG^o = ΔH^o–TΔS^o.

Temperature must be in K (Add 273 to °C to get temperature in K); so 20+273= 293K.

Units for enthalpy include kilojoules and units for entropy include joules. So it is best to convert entropy into kilojoules by dividing by 1000; +138 J K^–1 mol^–1 becomes +0.138 kJ K^–1 mol^–1.

Therefore ΔG^o = +96−(293×+0.138) = 96−40.434

= +55.566 = +55.6 kJ mol^–1 to three sig figs

Thus the correct answer is +55.6

(The standard free energy value is positive, so the reaction is not spontaneous. Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero.)

Incorrect answers

−40300 (−40338) is obtained if entropy value (+138) is not converted into kilojoules.

+93.2 (+93.24) is obtained if the temperature is not converted into Kelvin; 20 is used as a value instead of 293.

−2660 (−2664) is obtained if entropy value (+138) is not converted into kilojoules and the temperature is not converted into Kelvin; 20 is used as a value instead of 293.

The standard free energy change (ΔG^o) can be calculated by using ΔG^o = ΔH^o–TΔS^o.

Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero. So to find the point at which the reaction becomes spontaneous we must calculate the temperature at which ΔG^o = 0.

Units for enthalpy include kilojoules and units for entropy include joules. So it is best to convert entropy into kilojoules by dividing by 1000; +138 J K^–1 mol^–1 becomes +0.138 kJ K^–1 mol^–1.

Therefore ΔG^o = 0 = +96−(T×+0.138)

Rearranging:

−(T×+0.138) = −96

(T×+0.138) = 96

T = 96/0.138

= 695.652174 = 696K to three sig figs

Thus the correct answer is 696

Incorrect answers

0.696 is obtained if entropy value (+138) is not converted into kilojoules.

54.9 (54.876) is the ΔG^o value (in kJ mol^–1) obtained if the values given in the question are plugged into ΔG^o = ΔH^o–TΔS^o (at 298K).

13.2 (13.248) is obtained if T = 96×0.138 instead of 96÷0.138 (rearranging error).