Question

Consider a reaction :
A(g) + B(g) $\rightleftharpoons$ C(g)
The backward reaction is favoured by :

• A. Decrease in pressure
• B. Increase in pressure
• C. Increase in concentration of $\left[B\right]$
• D. None of the above

Answer 1

The correct option is A Decrease in pressure

According to the given reaction,
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)$

Number of moles of reactants are more than the number of moles of products. $\Delta n_g=1-2=-1\Delta n_g<0$
According to the Le - Chatelier's principle, on decreasing the pressure, reaction will tend to go in that direction where the number of moles are more i.e. $n_g\uparrow$
Number of moles are more in the reactant side i.e. backward direction.
So, backward direction is favoured by decrease in pressure.

Theory:

Effect of change in pressure:
At constant temperature when the pressure increases then, the ratio of $\frac{n_g}{V}$ (because of increase in population density) increases.
When, $\Delta n_g>0$
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$ When, $\Delta n_g=+ve$ then total moles increase from left to right as the reaction moves forward.
Example :
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$
At constant volume, if pressure increases then $\frac{n_g}{V}$ also increases. Reaction goes backward, that is total moles decrease from right to left.

Similarly when, $\Delta n_g<0$
consider $2A\left(g\right)\rightleftharpoons B\left(g\right)$
$\Delta n_g=-ve$ i.e. total moles decreases from left to right.
At constant volume, if pressure increases then $\frac{n_g}{V}$ also increases. Reaction goes forward.
When, $\Delta n_g=0$ consider $A\left(g\right)\rightleftharpoons B\left(g\right)$ When, $\Delta n_g=0$. Here even if pressure changes there is no shift in equilibrium.

note: Solids and liquids are incompressible or negligible. Therefore change in pressure has no effect on such equilibria.