Solveeit Logo
Login

Question

Question: For the equilibrium, \(A(g)\rightleftharpoons B(g),\Delta H=-40kJ/mol\) if the ratio of the activati...

For the equilibrium, A(g)⇌B(g),ΔH=−40kJ/molA(g)\rightleftharpoons B(g),\Delta H=-40kJ/mol if the ratio of the activation energies of the forward (Ef{{E}_{f}} ) and reverse ( Eb{{E}_{b}} ) reaction is 23\dfrac{2}{3} then:
(A) Ef=30kJ/mol;Eb=70kJ/mol{{E}_{f}}=30kJ/mol;{{E}_{b}}=70kJ/mol
(B) Ef=70kJ/mol;Eb=30kJ/mol{{E}_{f}}=70kJ/mol;{{E}_{b}}=30kJ/mol
(C) Ef=80kJ/mol;Eb=120kJ/mol{{E}_{f}}=80kJ/mol;{{E}_{b}}=120kJ/mol
(D) Ef=60kJ/mol;Eb=100kJ/mol{{E}_{f}}=60kJ/mol;{{E}_{b}}=100kJ/mol

Explanation

Solution

If the temperature of the reaction, the rate of reaction increases based on collision theory and Maxwell Boltzmann distribution of molecular energies in a gas. Collisions in a reaction result in the particles colloid with enough energy to get the reaction started, by which the minimum energy required is called activation energy.

Complete step by step solution:
Because of increasing temperature reaction rate increases and a disproportionately large number of high collisions possessing at least the activation energy of the reaction. only these collisions result in a reaction due to their activation energy.
Given equilibrium reaction,
A(g)⇌B(g),ΔH=−40kJ/molA(g)\rightleftharpoons B(g),\Delta H=-40kJ/mol
The above reaction is exothermic because the change in enthalpy shows a negative sign.
The change in enthalpy is equal to the difference between activation energies of forwarding reaction and backward reaction.
ΔH=Ef−Eb\Delta H={{E}_{f}}-{{E}_{b}} --- (1)
Given the ratio of activation energies of forward and backward reaction is,
EfEb=23\dfrac{{{E}_{f}}}{{{E}_{b}}}=\dfrac{2}{3}
Then, Ef=23Eb{{E}_{f}}=\dfrac{2}{3}{{E}_{b}} --- (2)
Substitute the above value in equation (1), and ΔH=−40kJ/mol\Delta H=-40kJ/mol
−40=Eb−23Eb-40={{E}_{b}}-\dfrac{2}{3}{{E}_{b}}
Eb=−120kJ/mol{{E}_{b}}=-120kJ/mol
Substitute the value of activation energy of a backward reaction in equation (2),
Ef=23Eb=23X(−120)=−80kJ/mol{{E}_{f}}=\dfrac{2}{3}{{E}_{b}}=\dfrac{2}{3}X(-120)=-80kJ/mol
Hence, Ef=80kJ/mol;Eb=120kJ/mol{{E}_{f}}=80kJ/mol;{{E}_{b}}=120kJ/mol

The correct answer is option C.

Note: The fraction of molecules present in a gas which have energies equal to or above activation energy at a particular temperature. At an absolute temperature, the fraction of molecules that have kinetic energy than the activation energy. The activation energy of a chemical reaction can be calculated by the Arrhenius equation.