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  • Rate constant of a reaction is given by: <img alt="\mathrm{k=2.4 \times 10^{12} e^{-\frac{15000}{R T}} -1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bk%3D2.4%2

Rate constant of a reaction is given by:

\mathrm{k=2.4 \times 10^{12} e^{-\frac{15000}{R T}} -1}. It means 

 

Option: 1

\mathrm{\log k ~V s~ \log T} with give a straight line of slopes 5000 .
 


Option: 2

\mathrm{\log k ~V s^{1 / T}} willgire a straight line of slope -15000 .
 


Option: 3

\mathrm{\log k~ vs~ \log T} willie a straight line of slope -15000 .
 


Option: 4

\mathrm{ \log k~ vs~ \frac{1}{T}} will give a straight line.
 


Answers (1)

best_answer

\begin{aligned} &\mathrm{ \log k=\log \left(2.4 \times 10^{12}\right)-\frac{15000}{2.303 R T} }\\ &\mathrm{ \log k \text { V } \frac{1}{T} \text { Graph } \rightarrow \text { straight line } }\\ &\mathrm{ \text { slope }=-\frac{15000}{2.303 R}}\\ \end{aligned}

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