Medium
The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.
- A$2.303 \times 2$
- B$2/2.303$
- C$2$
- DNone of these
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Similar Questions
The half-life periods of a first order reaction at $300 \ K$ and $400 \ K$ are $50 \ s$ and $10 \ s$ respectively. The activation energy of the reaction in $kJ \ mol^{-1}$ is $(\log 5 = 0.70)$
The rate of a certain reaction doubles when the temperature changes from $27\,^oC$ to $37\,^oC$. The activation energy for the reaction is ........... $kJ\,mol^{-1}$.
Difficult
View Solution$A$ large increase in the rate of reaction for a small increase in temperature is due to:
Easy
View SolutionCatalyst $A$ reduces the activation energy for a reaction by $10 \ kJ \ mol^{-1}$ at $300 \ K$. The ratio of rate $\frac{k_{T, \text{Catalysed}}}{k_{T, \text{Uncatalysed}}}$ is $e^{x}$. Find the value of $x$ [nearest integer].
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]
DifficultJEE MAIN 2022
View SolutionThe rate constant,$k$ for a first order reaction,$C_2H_5I_{(g)} \rightarrow C_2H_{4(g)} + HI_{(g)}$ is $x \ s^{-1}$ at $600 \ K$ and $4x \ s^{-1}$ at $700 \ K$. The energy of activation of the reaction (in $kJ \ mol^{-1}$) is (in $.16$)
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