Arrhenius equation and activation energy — practice questions

Question 1 · difficulty L2 · understanding

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer) Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Question 2 · difficulty L2 · understanding

For a reaction, given below is the graph of $$\ln k$$ vs $${1 \over T}$$. The activation energy for the reaction is equal to ____________ $$\mathrm{cal} \,\mathrm{mol}^{-1}$$. (nearest integer) (Given : $$\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}$$ )

Question 3 · difficulty L2 · understanding

The equation k = (6.5 $$\times$$ 10 12 s $$-$$1 )e $$-$$26000K/T is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol $$-$$1 . [nearest integer] (Given : R = 8.314 J K $$-$$1 mol $$-$$1 )

Question 4 · difficulty L2 · understanding

The rate constant for a first order reaction is given by the following equation: $$\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T}$$ The activation energy for the reaction is given by ____________ kJ mol $$-$$1 . (In nearest integer) (Given : R = 8.3 J K $$-$$1 mol $$-$$1 )

Question 5 · difficulty L2 · understanding

If the activation energy of a reaction is 80.9 kJ mol $$-$$1 , the fraction of molecules at 700 K, having enough energy to react to form products is e $$-$$x . The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K $$-$$1 mol $$-$$1 ]

Question 6 · difficulty L2 · understanding

Consider the given figure and choose the correct option :

• A. Activation energy of both forward and backward reaction is $E_1+E_2$ and reactant is more stable than product.
• B. Activation energy of forward reaction is $E_1+E_2$ and product is more stable than reactant.
• C. Activation energy of backward reaction is $\mathrm{E}_1$ and product is more stable than reactant.
• D. Activation energy of forward reaction is $E_1+E_2$ and product is less stable than reactant.

Question 7 · difficulty L2 · understanding

For a first-order reaction A $$\to$$ P, the temperature (T) dependent rate constant (k) was found to follow the equation $$\log k = - (2000){1 \over T} + 6.0$$. The pre-exponential factor A and activation energy $$E_a$$, respectively, are

• A. $$1.0\times10^6~\mathrm{s^{-1}}$$ and 9.2 kJ mol$$^{-1}$$
• B. $$6.0~\mathrm{s^{-1}}$$ and 16.6 kJ mol$$^{-1}$$
• C. $$1.0\times10^6~\mathrm{s^{-1}}$$ and 16.6 kJ mol$$^{-1}$$
• D. $$1.0\times10^6~\mathrm{s^{-1}}$$ and 38.3 kJ mol$$^{-1}$$

Question 8 · difficulty L3 · understanding

For reaction A → P, rate constant $k = 1.5 \times 10^3 \ \mathrm{s}^{-1}$ at $27^{\circ}\mathrm{C}$ If activation energy for the above reaction is $60\ \mathrm{kJ}\ \mathrm{mol}^{-1}$, then the temperature (in $^{\circ}\mathrm{C}$) at which rate constant, $k = 4.5 \times 10^3\ \mathrm{s}^{-1}$ is ______. (Nearest integer) Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3\ \mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}$, $\ln 10 = 2.3$

Question 9 · difficulty L3 · understanding

Consider $\mathrm{A} \xrightarrow{\mathrm{k}_1} \mathrm{~B}$ and $\mathrm{C} \xrightarrow{\mathrm{k}_2} \mathrm{D}$ are two reactions. If the rate constant $\left(\mathrm{k}_1\right)$ of the $\mathrm{A} \longrightarrow \mathrm{B}$ reaction can be expressed by the following equation $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$ and activation energy of $C \longrightarrow D$ reaction $\left(E a_2\right)$ is $\frac{1}{5}$ th of the $A \longrightarrow B$ reaction $\left(E a_1\right)$, then the value of $\left(E a_2\right)$ is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Question 10 · difficulty L3 · understanding

The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer) $$\begin{array}{ll} \mathrm{X} \longrightarrow \mathrm{Y}, & \mathrm{k}_1=10^6 e^{\frac{-30000}{\mathrm{~T}}} \\ \mathrm{P} \longrightarrow \mathrm{Q}, & \mathrm{k}_2=10^4 e^{\frac{-24000}{\mathrm{~T}}} \end{array}$$ Given : $\ln 10=2.303$