Question

For which of the following reactions at equilibrium at constant temperature, doubling the volume will shift equilibrium towards right?

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
  
• B. $CaC{O}_3\left(s\right)\rightleftharpoons Ca{O}_{(}s)+C{O}_2(g)$
  
• C. $2CO\left(g\right)+O_2\rightleftharpoons 2C{O}_2\left(g\right)$
  
• D. $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$

Answer 1

Answer: (A), (B)

$CaC{O}_3\left(s\right)\rightleftharpoons Ca{O}_{(}s)+C{O}_2(g)$


Calculating $\Delta n_g$
Option A : $2-1=1=+ve$
Option B : $1-0=1=+ve$
Option C : $2-3=-1=-ve$
Option D : $2-2=0$

Doubling the volume would reduce the pressure and shift the equilibrium in the direction where there is more number of moles of gases i.e where $\Delta n_g$is $+ve$
In choices (A) and (B), $\Delta n_g$ is $+ve$, and thus on doubling volume, the equilibrium will shift to the right.