Question

Consider the equilibrium set up:
$2H_{2\left(g\right)}+C{O}_{\left(g\right)}\rightleftharpoons C{H}_{3}O{H}_{\left(g\right)}$
What will be the effect of the following on the equilibrium of the reaction?
(i) Addition of $H_2$
(ii) Addition of $C{H}_{3}OH$
(iii) Removal of CO
(iv) Removal of $C{H}_{3}OH$

• A. (i) - Forward direction (ii) - Backward direction (iii) - Backward direction (iv) - Forward direction
• B. (i) - Backward direction (ii) - Backward direction (iii) - Forward direction (iv) - Forward direction
• C. (i) - Forward direction (ii) - Forward direction (iii) - Backward direction (iv) - Backward direction
• D. (i) - Backward direction (ii) - Forward direction (iii) - Forward direction (iv) - Backward direction

Answer 1

The correct option is A (i) - Forward direction (ii) - Backward direction (iii) - Backward direction (iv) - Forward direction

$2H_2\left(g\right)+CO\left(g\right)\rightleftharpoons C{H}_{3}OH$

(i) Addition of Hydrogen favours forward reaction, as the number of moles of reactants increases, reaction shifts towards right side.

(ii) Addition of $C{H}_{3}OH$ favours reverse reaction in order to compensate the increase in the concentration of the product.

(iii) Removal of CO favours backward reaction to increase the concentration of CO.

(iv) Removal of $C{H}_{3}OH$ favours forward reaction (as the concentration decreases in one end the reaction proceeds towards the other end).