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,

When $\Delta n_g>0$ and $P\uparrow$, the reaction goes in backward direction.

Also, when $\Delta n_g<0$, and $P\uparrow$, the reaction goes in forward direction.


Clearly we can see that, when $\Delta n_g>0$, the increase in pressure cannot increase the yield of products to be formed as reaction goes in backward direction.

Calculating $\Delta n_g$ in all of the reactions:
a): $\Delta n_g=2-3=-1$
b): $\Delta n_g=2-4=-2$
c): $\Delta n_g=2-1=1$


In option a and b both, $\Delta n_g<0$ and in option C, $\Delta n_g>0$

So, yield of product cannot be increased in option c) by application of high pressure.