Question

Reaction in which, the yield of product(s) increases with an increase in pressure is:

• A. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• B. $H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$
• C. $H_{2}O\left(g\right)+C\left(s\right)\rightleftharpoons CO\left(g\right)+H_2\left(g\right)$
• D. $CO\left(g\right)+3H_2\left(g\right)\rightleftharpoons C{H}_4\left(g\right)+H_{2}O\left(g\right)$

Answer 1

The correct option is D $CO\left(g\right)+3H_2\left(g\right)\rightleftharpoons C{H}_4\left(g\right)+H_{2}O\left(g\right)$

According to Le-Chatelier's principle :
At equilibrium , if the pressure increases then equilibrium will shift towards a side having lesser number of moles (gaseous) so that the effect of change can be minimized.

• In option (a) and option (b) as the difference in number of gaseous moles in products and reactants side is zero i.e. $\Delta n_g=0$. So, no effect of pressure is there on equilibrium.
• In option (c) , the gaseous moles on the reactant side is 1 and on the product side are 2. So, $\Delta n_g=(2-1)=1$ . From reactants to products, the number of gaseous moles are increasing. Therefore, equilibrium shifts in backward direction.
• In the equilibrium reaction (d), $\Delta n_g=(2-4)=-2$ . From reactant to products, the number of gaseous moles decreases. Therefore, on increasing pressure, the equilibrium will shift in forward direction and the yield of products will increase.