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It has been found that for a chemical reaction with rise in temperature by 9 K  the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation enegy is found to be _________\mathrm{kJ}\; \mathrm{mol}^{-1} \text {. }\;[nearest integer]

\text { (Given } \left.\ln 10=2.3, \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 2=0.30\right)

Option: 1

59


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} &\mathrm{K_{300}=A \exp \left(\frac{-E a}{300 R}\right)} \\ &\mathrm{K_{309}=A \exp \left(\frac{-E a}{309 R}\right)} \end{aligned}

Given,

\mathrm{K_{309}=2 \; K_{300}}

\mathrm{\exp \left(\frac{-E a}{309 R}\right)=2 \exp \left(\frac{-E a}{300 R}\right)}

\mathrm{\frac{-E a}{309 R}=\ln 2-\frac{E a}{300 R}}

\mathrm{\Rightarrow \frac{E a}{R}\left(\frac{1}{300}-\frac{1}{309}\right)=\ln 2}

\mathrm{\Rightarrow \quad Ea=\frac{300 \times 309 \times \ln 2 \times R}{9}}

\begin{aligned} \Rightarrow\mathrm{ \quad Ea =58988.1 ~J} \\ \end{aligned}

                   \begin{aligned} &=58.9 \mathrm{~kJ} \end{aligned}

Hence, answer is 59

Posted by

Anam Khan

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