The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value ($R=$ gas constant,$E_a=$ Activation energy)
- A$\frac{R}{E_a}$
- B$\frac{E_a}{R}$
- C$\frac{-E_a}{R}$
- D$\frac{-R}{E_a}$
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Similar Questions
For a reaction,the rate constants $K_1$ and $K_2$ are given by $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T}$ respectively. At what temperature will $K_1 = K_2$?
Difficult
View SolutionThe rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is
MediumKCET 2023
View SolutionSubtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.
Easy
View SolutionThe rate of a reaction:
Easy
View SolutionThe rate of a reaction doubles,when the temperature is changed from $300 \ K$ to $310 \ K$. Activation energy of the reaction is....... $(R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 2=0.301)$
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