Reactivity 2.2.12—The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. Describe the qualitative relationship between temperature and the rate constant. Analyse graphical representations of the Arrhenius equation, including its linear form.

Which substance is the reducing agent in the given reaction?

H⁺ (aq) + 2H₂O (l) + 2MnO₄⁻ (aq) + 5SO₂ (g) → 2Mn²⁺ (aq) + 5HSO₄⁻ (aq)

A.  H⁺

B.  H₂O

C.  MnO₄⁻

D.  SO₂

Which substance is the reducing agent in the given reaction?

H⁺ (aq) + 2H₂O (l) + 2MnO₄⁻ (aq) + 5SO₂ (g) → 2Mn²⁺ (aq) + 5HSO₄⁻ (aq)

A.  H⁺

B.  H₂O

C.  MnO₄⁻

D.  SO₂

Which graph represents the relationship between the rate constant, $k$, and temperature, $T$, in kelvin?

Which graph represents the relationship between the rate constant, $k$, and temperature, $T$, in kelvin?

Which graph represents the relationship between the rate constant, $k$, and temperature, $T$, in kelvin?

Which graph represents the relationship between the rate constant, $k$, and temperature, $T$, in kelvin?

What is the activation energy according to the following plot of the linear form of the Arrhenius equation?

Arrhenius equation: $k=A{e}^{\frac{\mathit{-}Ea}{RT}}$.

A.  $E_a=\frac{2\times 8.31}{0.06}$

B.  $E_a=\frac{-2\times 8.31}{0.06}$

C.  $E_a=e^{\frac{2\times 8.31}{0.06}}$

D.  $E_a=e^{\frac{-2\times 8.31}{0.06}}$

What is the activation energy according to the following plot of the linear form of the Arrhenius equation?

Arrhenius equation: $k=A{e}^{\frac{\mathit{-}Ea}{RT}}$.

A.  $E_a=\frac{2\times 8.31}{0.06}$

B.  $E_a=\frac{-2\times 8.31}{0.06}$

C.  $E_a=e^{\frac{2\times 8.31}{0.06}}$

D.  $E_a=e^{\frac{-2\times 8.31}{0.06}}$

What is the activation energy according to the following plot of the linear form of the Arrhenius equation?

Arrhenius equation: $k=A{e}^{\frac{\mathit{-}Ea}{RT}}$.

A.  $E_a=\frac{2\times 8.31}{0.06}$

B.  $E_a=\frac{-2\times 8.31}{0.06}$

C.  $E_a=e^{\frac{2\times 8.31}{0.06}}$

D.  $E_a=e^{\frac{-2\times 8.31}{0.06}}$

What is the activation energy according to the following plot of the linear form of the Arrhenius equation?

Arrhenius equation: $k=A{e}^{\frac{\mathit{-}Ea}{RT}}$.

A.  $E_a=\frac{2\times 8.31}{0.06}$

B.  $E_a=\frac{-2\times 8.31}{0.06}$

C.  $E_a=e^{\frac{2\times 8.31}{0.06}}$

D.  $E_a=e^{\frac{-2\times 8.31}{0.06}}$

The activation energy of a reaction can be obtained from the rate constant, k, and the absolute temperature, $\text{T}$. Which graph of these quantities produces a straight line?

A.  k against $\text{T}$

B.  k against $\frac{1}{\text{T}}$

C.  ln k against $\text{T}$

D.  ln k against $\frac{1}{\text{T}}$

The activation energy of a reaction can be obtained from the rate constant, k, and the absolute temperature, $\text{T}$. Which graph of these quantities produces a straight line?

A.  k against $\text{T}$

B.  k against $\frac{1}{\text{T}}$

C.  ln k against $\text{T}$

D.  ln k against $\frac{1}{\text{T}}$

The activation energy of a reaction can be obtained from the rate constant, k, and the absolute temperature, $\text{T}$. Which graph of these quantities produces a straight line?

A.  k against $\text{T}$

B.  k against $\frac{1}{\text{T}}$

C.  ln k against $\text{T}$

D.  ln k against $\frac{1}{\text{T}}$

The activation energy of a reaction can be obtained from the rate constant, k, and the absolute temperature, $\text{T}$. Which graph of these quantities produces a straight line?

A.  k against $\text{T}$

B.  k against $\frac{1}{\text{T}}$

C.  ln k against $\text{T}$

D.  ln k against $\frac{1}{\text{T}}$