For the reaction X rightleftharpoons Y,the pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to the Arrhenius equation,the rate constant $k$ is given by $k = A e^{-E_a/RT}$.For the forward reaction: $k_f = A_f e^{-E_{af}/RT}$.For

Vedclass · Accepted Answer

$e^{-\Delta E/RT}$

Vedclass · Answer

(C) According to the Arrhenius equation,the rate constant $k$ is given by $k = A e^{-E_a/RT}$.For the forward reaction: $k_f = A_f e^{-E_{af}/RT}$.For the backward reaction: $k_b = A_b e^{-E_{ab}/RT}$.Given that the pre-exponential factors are equal,$A_f = A_b = A$.The equilibrium constant $K_c$ is defined as $K_c = \frac{k_f}{k_b}$.Substituting the expressions: $K_c = \frac{A e^{-E_{af}/RT}}{A e^{-E_{ab}/RT}} = e^{(E_{ab} - E_{af})/RT}$.Since $\Delta E = E_{af} - E_{ab}$,it follows that $E_{ab} - E_{af} = -\Delta E$.Therefore,$K_c = e^{-\Delta E/RT}$.

For the reaction $X \rightleftharpoons Y$,the pre-exponential factors for the forward and backward reactions are equal. What will be the equilibrium constant of the reaction?

Similar Questions

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

$A$ large increase in the rate of a reaction for a rise in temperature is due to:

The addition of a catalyst during a chemical reaction alters which of the following quantities?

Activation energy of any chemical reaction can be calculated if one knows the value of

The minimum energy a molecule should possess in order to enter into a fruitful collision is known as

Vedclass Test Series

Exam Paper Generator

Online Exam Module