Reaction Quotient Q And Gibbs Energy G

THE REACTION QUOTIENT 'Q'

Consider the equilibrium

PCl₅ (g) $\rightleftharpoons$ PCl₃(g) + Cl₂ (g)

At equilibrium $\frac{\left[C{l}_2\right]\left[PC{l}_3\right]}{\left[PC{l}_5\right]}$ = K_C. When the reaction is not at equilibrium this ratio is called 'Q_C' i.e., Q_C is the general term used for the above given ratio at any instant of time. And at equilibrium Q_C becomes K_C.

Similarly, $\frac{{\text{P}}_{\text{C}{\text{l}}_{\text{2}}}{\text{P}}_{\text{PC}{\text{l}}_{\text{3}}}}{{\text{P}}_{\text{PC}{\text{l}}_{\text{5}}}}$is called Q_P and at equilibrium it becomes K_P.

$\bullet$ If the reaction is at equilibrium, Q = K_c

$\bullet$ A net reaction proceeds from left to right (forward direction) if Q < K_C.

$\bullet$ A net reaction proceeds from right to left (the reverse direction) if Q >K_c

Illustration 1. For the reaction,

A_(g) + B_(g) $\rightleftharpoons$ 2C_(g) at 25°C, in a 2 litre vessel contains 1, 2, 3 moles of respectively. Predict the direction of the reaction if

(a) K­c for the reaction is 3

(b) Kc for the reaction is 6

(c) Kc for the reaction is 4.5

Solution: A_(g) + B_(g) $\rightleftharpoons$ 2C_(g)

Reaction quotient Q = $\frac{{[C]}^2}{[A][B]}=\frac{{\left(\frac{3}{2}\right)}^2}{\frac{1}{2}\times \frac{2}{2}}=\frac{9}{2}=4.5$

(a) $\therefore Q>K_c,$therefore backward reaction will be followed

(b) $\because Q<K_c$

$\because$The forward reaction is followed

(c) Q = K_c

$\because$ The reaction is at equilibrium

Standard Free Energy Change of a Reaction and its Equilibrium Constant

Let $\Delta$G⁰ be the difference in free energy of the reaction when all the reactants and products are in the standard state and K_c or, K_p be the thermodynamic equilibrium constant of the reaction. Both related to each other at temperature T by the following relation.

$\Delta$G⁰ = – 2.303 RT logK_c and $\Rightarrow$G⁰ = – 2.303 RT log K_p (in case of ideal gases)

Now, we know that thermodynamically,

$\Delta$G⁰ = $\Delta$H⁰ – T$\Delta$S⁰

here $\Delta$H⁰ is standard enthalpy of reaction, and $\Delta$S⁰ is standard entropy change

(i) When $\Delta$G⁰ = 0, then, K_c = 1

(ii) When, $\Delta$G⁰ > 0, i.e. +ve, then K_c < 1, in this case reverse reaction is feasible showing thereby a less concentration of products at equilibrium rate.

(iii) When $\Delta$G⁰ < 0, i.e. -ve, then, K_c > 1; In this case, forward reaction is feasible showing thereby a large concentrations of product at equilibrium state.

Illustration 2. A system is in equilibrium as

PCl₅ + Heat $\rightleftharpoons$ PCl₃ + Cl₂

Why does the temperature of the system decreases, when PCl₃ are being removed from the equilibrium mixture at constant volume?

Solution: When PCl₃ are being removed from the system, the reaction moved to the right. This consumed heat and therefore, temperature is decreased.