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For the equilibrium 2NO2 (g)⇌N2O4 (g)+ 14.6 kcal the increase in temperature would
A. Favour the formation of N2O4
B. Favour the decomposition of N2O4
C. Does not alter the equilibrium
D. Stop the reaction

Answer
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Hint: The rate of reaction is affected by increasing or decreasing pressure and temperature. In an endothermic reversible reaction, the temperature increases the rate of forward reaction (i.e, in favour of the reactant). In an exothermic reversible reaction, the temperature increases the rate of backward reaction (i.e, in favour of products).

Complete answer:A chemical reaction in which the formation of product and as well as the formation of reactant takes place such type of reaction is called a reversible reaction. And reversible reactions try to attain chemical equilibrium. Equilibrium is a state where the rate of formation of product is equal to the rate of formation of reactant. As we know endothermic reactions absorb the heat energy/temperature in the reaction, and exothermic reaction releases the heat energy/temperature in the reaction.
In the given reaction, 2NO2 (g)⇌N2O4 (g)+ 14.6 kcal we can see that in the forward reaction NO2 converts into N2O4 and releases 14.6 kcal heat energy in the reaction, it is an exothermic reaction. In the backward reaction N2O4 converts into NO2 and absorbs 14.6 kcal heat energy in the reaction, it is an endothermic reaction. Thus, on the increasing temperature, the rate of backward reaction will increase i.e, conversion of N2O4 into NO2. Hence, increase in temperature favours the decomposition of N2O4.

Thus, Option (B) is correct

Note: The reversible reaction is affected by pressure. If the reactant moles are more in the reaction then on increasing pressure the rate of forward reaction will increase and backward reaction will decrease and on decreasing pressure the rate of backward reaction will increase and forward reaction will increase.