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Consider the following reaction : \mathrm{N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g);\; \Delta H^{0}=+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 N2 at constant T.
Option: 1 (a) towards product, (b) towards reactant  
Option: 2 (a) towards reactant, (b) towards product
Option: 3 (a) towards reactant, (b) no change
Option: 4 (a) towards product, (b) no change

Answers (1)

best_answer

N2(g) + 3H2(g) ? 2NH2(g) + heat (exothermic)

(a) Temperature decreased → favors product (forward reaction)
(b) Adding N2 at constant T → increases reactant → shifts towards product

Correct Answer: Option 4 — (a) towards product, (b) no change

If N? is added as a reactant, then Option 1 is correct.
So, the final answer depends on how pressure is increased.

Posted by

Divya Sharma

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