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Question: \({A_2}\left( g \right) \to 2A\left( g \right)\) and for this reaction on increasing \(T\) value of ...

A2(g)→2A(g){A_2}\left( g \right) \to 2A\left( g \right) and for this reaction on increasing TT value of Keq{K_{eq}} increases, then for this reaction is:
A.ΔH=\Delta H = positive; ΔS=\Delta S = positive
B.ΔH=\Delta H = negative; ΔS=\Delta S = negative
C.ΔH=\Delta H = positive; ΔS=\Delta S = negative
D.ΔH=\Delta H = negative; ΔS=\Delta S = positive

Explanation

Solution

To answer this question, you must recall the relation between the equilibrium constant of a reaction and its free energy. Gibbs free energy or the Gibbs function is a thermodynamic function which is used for the calculation of the maximum work performed by a system at constant temperature and pressure.
Formulae used: ΔGθ=−RTln⁡Keq\Delta {G^\theta } = - RT\ln {K_{eq}}
Where, ΔG\Delta G represents the change in the free energy during the reaction
Keq{K_{eq}} represents the equilibrium constant of the reaction.
And, TT represents the temperature of the reaction mixture.
The free energy gives a relation between the enthalpy change in a process and the change in entropy during a process as ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Complete step by step answer:
We know that the equilibrium constant of a reaction depends on the temperature of the reaction. In this case, the equilibrium constant is given to increase as we increase the temperature. So we can write the relation between the equilibrium constants at two different temperatures as
ln⁡(Keq)2(Keq)1=−ΔGR[1T2−1T1]\ln \dfrac{{{{\left( {{K_{eq}}} \right)}_2}}}{{{{\left( {{K_{eq}}} \right)}_1}}} = - \dfrac{{\Delta G}}{R}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]
Since we know that, for T2>T1{T_2} > {T_1}the equilibrium constant (Keq)2>(Keq)1{\left( {{K_{eq}}} \right)_2} > {\left( {{K_{eq}}} \right)_1}
So, from this, we can say that the change in the Gibbs free energy of the reaction is positive, i.e. ΔG>0\Delta G > 0
The reaction occurring in the question is A2(g)→2A(g){A_2}\left( g \right) \to 2A\left( g \right). We can see that the number of moles of gases increases and thus the entropy will increase. Hence the change in entropy will be positive.
⇒ΔS=\Rightarrow \Delta S =positive.
Thus, ΔH=\Delta H = positive.

Thus, the correct answer is A.
Note:
The Gibbs free energy gives a relation between the enthalpy change in a process and the change in entropy during a process as ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.
In the above question, the change in the Gibbs free energy is positive and the entropy is also positive. Thus, the enthalpy change must be positive in order to maintain the equality.