Consider the following reaction:
\(\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}\); \(\Delta H^\circ = +58\) kJ
For each of the following cases (a, b), the direction in which the equilibrium shifts is :
(a) Temperature is decreased.
(b) Pressure is increased by adding \(\mathrm{N_2}\) at constant T.
\(\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}\); \(\Delta H^\circ = +58\) kJ
For each of the following cases (a, b), the direction in which the equilibrium shifts is :
(a) Temperature is decreased.
(b) Pressure is increased by adding \(\mathrm{N_2}\) at constant T.
Show Hint
- (a) towards reactant, (b) towards product
- (a) towards reactant, (b) no change
- (a) towards product, (b) towards reactant
- (a) towards product, (b) no change
The Correct Option is B
Solution and Explanation
(a) Temperature is decreased: by Le Chatelier's principle the system opposes the cooling by favouring the exothermic (reverse) reaction, so the equilibrium shifts towards the reactant (\(\mathrm{N_2O_4}\)).
(b) \(\mathrm{N_2}\) is added at constant T (constant volume). Total pressure rises but the partial pressures (concentrations) of \(\mathrm{N_2O_4}\) and \(\mathrm{NO_2}\) are unchanged, so \(Q_c = K_c\) still holds and there is no shift.
Hence (a) towards reactant, (b) no change.
Answer: (2) (a) towards reactant, (b) no change
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