Question

The reaction in which yield of the products cannot be increased by the application of high pressure is:

• A. $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
• B. $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
• C. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
• D. None of the above

Answer 1

The correct option is C $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

According to the Le- chatelier's principle,
increase in pressure favours the reaction to move in the direction where the number of moles are less.
In the following two reactions,
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
the number of moles of reactants is more than the product, so it will result in increase in yield whereas in
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ ,
the number of moles of product is more than the reactant. Thus, we can say that by applying pressure, backward reaction is favoured and product yeild will be reduced.