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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
Write a relation between $\Delta\text{G}$ and Q and define the meaning of each term and answer the following:
  1. Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K?
  2. Explain the effect of increase in pressure in terms of reaction quotient Q.
For the reaction,
✓

Answer

The relation between $\Delta\text{G}$ and Q is
$\Delta\text{G}=\Delta\text{G}^{\ominus}+\text{RT InQ}$
$\Delta\text{G}=$ change in free energy as the reaction proceeds
$\Delta\text{G}^{\ominus}$ standard free energy
Q = reaction quotienten
R= gas constant
T = absolute temperature in Kual
  1. Since, $\Delta\text{G}^{\ominus}=-\text{RT In K}$
$\therefore\Delta\text{G}=-\text{RT In K}+\text{RT In Q;}$
$\Delta\text{G}=\text{RT In}\frac{\text{Q}}{\text{K}}$
will be negative and the reaction proceeds in the forward direction. $\text{Q}=\text{K},\Delta\text{G}=0$ If reaction is in equilibrium and there is no net reaction.
  1. $\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$
$\text{K}_{\text{c}}=\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$
On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,
$\text{Q}_{\text{c}}=\frac{2[\text{CH}_4].2[\text{H}_2\text{O}]}{2[\text{CO}]\{2[\text{H}_2]\}^3}$
$\frac{1}{4}\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}=\frac{1}{4}\text{K}_{\text{c}}$
Therefore, $Q_c$ is less than $K_C$ so $Q_c$ will tend to increase to re-establish equilibrium and the reaction will go in forward direction.
$\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$

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