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Question: At a certain temperature and total pressure of \[1atm\] , Equilibrium mixture contains \[20%\] by vo...

At a certain temperature and total pressure of 1atm1atm , Equilibrium mixture contains 2020% by volume of AA atoms A2(g)⇆2A(g){{A}_{2\left( g \right)}}\underset{{}}{\leftrightarrows}2{{A}_{(g)}} , The Kp{{K}_{p}} of the reaction is:
A. 5×10−25\times {{10}^{-2}}
B. 2×10−32\times {{10}^{-3}}
C. 5×10−35\times {{10}^{-3}}
D. 4×10−44\times {{10}^{-4}}

Explanation

Solution

As we know that Gibbs free energy is the energy that should be supplied to a system and only a small portion of it is converted to useful work which is our free energy, entropy is the randomness of the system and the enthalpy is the amount of heat supplied at constant pressure.

Complete step by step answer:
We know that Gibbs free energy is the energy that should be supplied to a system and only a small portion of it is converted to useful work which is our free energy, entropy is the randomness of the system and the enthalpy is the amount of heat supplied at constant pressure and the equilibrium constant is that value where the reaction has no tendency to move forward or backward.
The equilibrium constant in terms of concentration KC{{K}_{C}} is defined as the ratio of the concentration of products to the concentration of the reactants. The equilibrium constant in terms of partial pressure Kp{{K}_{p}} is defined as the ratio of the partial pressure of products to the partial pressure of the reactants.
Total Pressure,P=1atm=105PaP=1atm={{10}^{5}}Pa
As 2020%molecules are AA atoms 8080%are A2{{A}_{2}} molecules.
Partial pressure of A(g),pA =0.2× 105PaA\left( g \right),{{p}_{A~}}=0.2\times ~{{10}^{5}}Pa
Partial pressure of A2(g),pA2=0.8× 105Pa{{A}_{2}}\left( g \right),{{p}_{A2}}=0.8\times ~{{10}^{5}}Pa
Kp=(pA)2pA2=(0.2×105)20.8×105=5×10−3{{K}_{p}}=\dfrac{{{({{p}_{A}})}^{2}}}{{{p}_{{{A}_{2}}}}}=\dfrac{{{(0.2\times {{10}^{5}})}^{2}}}{0.8\times {{10}^{5}}}=5\times {{10}^{-3}}
Therefore, the correct answer is option C, i.e. Kp{{K}_{p}} of the reaction is 5×10−35\times {{10}^{-3}}

Note: Always remember that for a reaction to be spontaneous, Gibbs free energy is always spontaneous when its value is present in negative, similarly change in enthalpy should also be negative and change in entropy should be always positive and for a nonspontaneous reaction to occur, Gibbs free energy is always positive, change in enthalpy is also positive and change in entropy is always negative. At equilibrium all these dimensions are zero.