Medium
The rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{\circ}C$,the activation energy of the reaction is (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 2 = 0.69$ and $\ln 3 = 1.1$)
- A$1.69 \ kJ \ mol^{-1}$
- B$0.169 \ kJ \ mol^{-1}$
- C$49.8 \ kJ \ mol^{-1}$
- D$498 \ J \ mol^{-1}$
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For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$
Medium
View SolutionThe rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)
$A$ reaction having equal activation energies for forward and reverse reaction has:
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View SolutionFor a reaction,the threshold energy is $75 \ kJ/mol$. If the internal energy of the reactants is $20 \ kJ/mol$,the activation energy (in $kJ/mol$) is
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Medium
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