Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
(N/A) Thermodynamic feasibility (indicated by a negative value of Gibbs free energy change,$\Delta G < 0$) does not determine the rate of a reaction. $A$ reaction may be thermodynamically spontaneous,but its rate can be extremely slow due to a high activation energy $(E_a)$.
For example,the conversion of diamond to graphite is thermodynamically feasible ($\Delta G$ is negative),but the reaction is kinetically very slow at room temperature because it possesses a very high activation energy.
For example,the conversion of diamond to graphite is thermodynamically feasible ($\Delta G$ is negative),but the reaction is kinetically very slow at room temperature because it possesses a very high activation energy.
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Similar Questions
$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:
$Step$ $Rate\ Constant\ /\ Activation\ energy$ $Step\ 1$ $k_1, E_{a_1} = 180\ kJ/mol$ $Step\ 2$ $k_2, E_{a_2} = 80\ kJ/mol$ $Step\ 3$ $k_3, E_{a_3} = 50\ kJ/mol$
| $Step$ | $Rate\ Constant\ /\ Activation\ energy$ |
| $Step\ 1$ | $k_1, E_{a_1} = 180\ kJ/mol$ |
| $Step\ 2$ | $k_2, E_{a_2} = 80\ kJ/mol$ |
| $Step\ 3$ | $k_3, E_{a_3} = 50\ kJ/mol$ |
Difficult
View SolutionWhich of the following statements is correct?
Medium
View SolutionThe activation energy for a reaction is $9.0 \, kcal/mol$. The increase in the rate constant when its temperature is increased from $298 \, K$ to $308 \, K$ is $......... \, \%$.
Difficult
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 constant,the activation energy and the Arrhenius parameter of a chemical reaction at $25\,^oC$ are $3.0 \times 10^{-4}\,s^{-1}$,$104.4\,kJ\,mol^{-1}$ and $6.0 \times 10^{14}\,s^{-1}$ respectively. The value of the rate constant as $T \to \infty$ is
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