JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a certain reaction consider the plot of \(lnk\) versus \(1/T\) given in the figure. If the rate constant of this reaction at \(400\,K\) is \(10^{-5}\,s^{-1},\) then the rate constant at \(500\,K\) is
- A \(10^{-6}\,s^{-1}\)
- B \(2\times 10^{-4}\,s^{-1}\)
- C \(10^{-4}\,s^{-1}\)
- D \(4\times 10^{-4}\,s^{-1}\)
Answer & Solution
Correct Answer
(C) \(10^{-4}\,s^{-1}\)
Step-by-step Solution
Detailed explanation
\(\ln \, = \,\ln \,A\, - \,\frac{{Ea}}{{RT}}\, = \,\ln \,A\, - \,\frac{{4606}}{T}\) \(\ln \,\left( {\frac{k}{{{{10}^{ - 5}}}}} \right) = \left( {\frac{{Ea}}{R}} \right)\, \times \,\frac{{500 - 400}}{{500 \times 400}}\)…
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