The activation energy of a first-order reacti | vedclass

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(B) Let the rate constant without a catalyst be $k_1$ and with a catalyst be $k_2$.Activation energy without catalyst $(E_1) = 30\,kJ\,mol^{-1}$.Activ

Vedclass · Accepted Answer

$11$

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(B) Let the rate constant without a catalyst be $k_1$ and with a catalyst be $k_2$.Activation energy without catalyst $(E_1) = 30\,kJ\,mol^{-1}$.Activation energy with catalyst $(E_2) = 24\,kJ\,mol^{-1}$.Temperature $(T) = 25 + 273 = 298\,K$.According to the Arrhenius equation:$\ln(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{RT}$$\log(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{2.303RT}$$\log(\frac{k_2}{k_1}) = \frac{(30 - 24) 	imes 10^3\,J\,mol^{-1}}{2.303 	imes 8.314\,J\,K^{-1}\,mol^{-1} 	imes 298\,K}$$\log(\frac{k_2}{k_1}) = \frac{6000}{5705.84} \approx 1.0515$$\frac{k_2}{k_1} = 	ext{antilog}(1.0515) \approx 11.26$Thus,the rate of reaction increases by approximately $11$ times.

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

Similar Questions

The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.

$A$ graph plotted between $\log \,K$ vs $\frac{1}{T}$ for calculating activation energy is shown by:

The rate constant for the decomposition of a hydrocarbon is given as $K = (4.5 \times 10^{11} \ s^{-1}) e^{-(28000 \ K) / T}$. The activation energy of the reaction (in $J/mol$) is:

Define the following terms:
$(a)$ Activation energy
$(b)$ Activated complex

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

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