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  • Plot between log K versus  for a reaction <img alt="\mathrm{A

Plot between log K versus \mathrm{\frac{1}{T}} for a reaction \mathrm{A \longrightarrow B}. Gives the slope of -1500 . What will be the \mathrm{E_a} for the reaction \mathrm{(in \; \mathrm{kJ} / \mathrm{mol} )}

Option: 1

25.3 kJ/mol


Option: 2

35.3 kJ/mol


Option: 3

30.3 kJ/ mol


Option: 4

45.3 kJ/mol


Answers (1)

best_answer

According to Arrehius Equation

\begin{aligned} \log K & =\log A-\frac{E_a}{2.303 R\left(\frac{1}{T}\right)} \\ \\\text { slope }= & \frac{-E_a}{2.303 R}=-1500 \\ \\E_a & =1500 \times 2.303 \times 8.314 \\ \\& =25.316 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Posted by

jitender.kumar

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