Question

For the decomposition of HI the following logarithmic plot is shown : [R = 1.98 cal/mol-K]

The activation energy of the reaction is about

• A. 45600 cal
• B. 13500 cal
• C. 24600 cal
• D. 32300 cal

Answer 1

Answer: (A)

45600 cal

Answer 2

$= - \frac{E_{a}}{2.303R} \times 10^{- 3} \times \frac{10^{3}}{T}cons\tan t$ thus, slope of

graph will be$- \frac{E_{a} \times 10^{- 3}}{2.303R} = - \frac{4}{0.4}$

$\Rightarrow E_{a} = 2.303 \times 1.98 \times 10^{4} = 45600cal$