Question

Which of the following reaction will not be affected by increasing the pressure?

• A. $PC{l}_3\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
• B. $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
• C. $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
• D. $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$

Answer 1

Answer: (B), (D)

The correct options are
B $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
D $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$

According to Le - chatelier's principle, if the number of moles of gas is the same on both sides of the reaction then the pressure has no effect i.e., reaction which have $\Delta n=0$ won't be affected by any change in pressure.
$\Delta n=\text{number of moles of gaseous product - number of moles of gaseous reactant}$
a.
$PC{l}_3\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ $\Delta n=1+1-1=1$
b.
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$ $\Delta n=2-(1+1)=0$
c.
$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$$\Delta n=1-0=1$
d. $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$$\Delta n=1+1-(1+1)=0$
$\Delta n$ is zero in (b) and (d), so (b) and (d) are not affected by pressure.