Write a relation between AG and Q and define the meaning of each term and answer the following:

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

(b) Explain the effect of an increase in pressure in terms of reaction quotient Q for the reaction:

CO (g) + 3H2(g) \rightleftharpoons CH4(g) + H2O (g)

Answers (1)

We can describe the relation between ΔG and Q as

ΔG = ΔG° + RT lnQ ...(1)

Herein, ΔG = Gibb’s energy change, ΔG° = standard Gibb’s energy, T = absolute temperature, Q = reaction quotient, and R = gas constant

We know that at Equilibrium, ΔG° = -RT lnK ...(2) (Wherein K is the Equilibrium constant)

Replacing the values of (2) in (1), we get

ΔG = -RT lnK + RT lnQ

ΔG = RT ln(Q/K) ...(3)

It is known to us that, Kc (which is the Equilibrium constant) is the ratio of concentration of products to that of reactants each raised to their stoichiometric coefficients at Equilibrium.

In contrast, Qc ( which is the reaction quotient) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at any time during the reaction.

(a) In this case, if Q< K, it implies that the concentration of products has to be increased in order to reach Equilibrium concentration. Hence, the net reaction will proceed in the forward direction.

When Q= K, it implies that the Equilibrium has been achieved and no net reaction occurs. Now lets answer tehs econd part of the question.

(b) CO (g) + 3H2(g) \rightleftharpoons CH4(g) + H2O (g) , As per the Le Chatiliers principle, when the pressure is increased, the Equilibrium will be shifting in that direction where there is a lesser number of gas molecules.

As per the available information,

Number of moles of reactants = 1 + 3 = 4

Number of moles of product = 1 + 1 = 2

Hence, the Equilibrium will shift in forward direction as number of moles of product is less. It is known to us that when Q < K, the Equilibrium will shift in the forward direction. Thus, for the given reaction Q < K

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