Medium
For an exothermic reaction $A \to B$,the activation energy is $65 \ kJ \ mol^{-1}$ and the heat of reaction is $-42 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction $B \to A$ would be....... $kJ \ mol^{-1}$
- A$23$
- B$107$
- C$65$
- D$42$
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Similar Questions
At $T \ K$,the following equation is obtained for a first order reaction: $\log \frac{k}{A} = -\frac{x}{T}$. The activation energy for this reaction is equal to $(R = \text{gas constant})$
$A$ large increase in the rate of reaction for a small increase in temperature is due to:
Easy
View SolutionExplain the effect of an increase in temperature on the rate of reaction and the rate constant.
Difficult
View SolutionThe rate of a first order reaction doubles when the temperature changes from $300 \ K$ to $310 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is
$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)
$R=8.3 \ J \ K^{-1} \ mol^{-1}, \log 2=0.3$ (in $.33$)
The rate constant for a reaction can be increased by $\underline{a}$ the stability of the reactant or by $\underline{b}$ the stability of the transition state. Select the correct choice for $a$ and $b$.
Medium
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