Question

Which is the correct representation at equilibrium?
(A) \[p = \dfrac{{eRT}}{N}\]
(B) \[k = {e^{ - \Delta {G^o}/RT}}\]
(C) \[\dfrac{{{K_1}}}{{{K_2}}} = {e^{ - {E_a}/RT}}\]
(D) \[\dfrac{p}{{{p^o}}} = {e^{ - \Delta H/RT}}\]

Answer 1

Hint: Gibbs free energy is the quantity used to measure the amount of work done in the system. It relates the standard Gibbs free energy, temperature, and universal gas constant. The Arrhenius equation sometimes relates the activation energy by the formula\[k = A{e^{ - {E_a}/RT}}\].

Complete Step by Step Solution:
Ideal gas law relates the terms pressure, temperature, moles, volume, and the universal gas constant. The formula of ideal gas law can be written as \[PV = nRT\], it can also be written as
\[P = \dfrac{{nRT}}{V}\]. Hence, an option is an incorrect answer and does not represent the equilibrium condition.

The Gibbs free energy can be written as \[\Delta G = \Delta {G^0} + RT\ln K\]
At equilibrium, the change in Gibbs energy will be zero. Hence, the above equation can be written as \[\Delta {G^0} = - RT\ln K\]
By writing in the exponential form, the equation can be written as \[k = {e^{ - \Delta {G^o}/RT}}\]
Hence, option b represents the condition at equilibrium.

The activation energy is the amount of energy required to cross the energy barrier. It can be denoted by \[{E_a}\]. The Arrhenius equation that relates the activation energy can be written as\[k = A{e^{ - {E_a}/RT}}\]. When this equation is written algebraically by taking two rate constants, it can be written as \[\ln \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]\].
By writing the above equation in exponential form, it will be \[\dfrac{{{K_1}}}{{{K_2}}} = {e^{ - {E_a}/RT}}\]
Hence, option c also represents the equilibrium condition.

The change in enthalpy can be written as \[\Delta H = RT\ln \left( {\dfrac{p}{{{p_o}}}} \right)\]
Hence, the given options D is an incorrect answer.

Note: At equilibrium, the term change in Gibbs energy will be zero, \[\Delta G = 0\]. It should be considered then only the Gibbs free energy equation can be written as option b. The Arrhenius equation should also be considered that was written in two rate constants.