Question

In which of the following equilibrium equation, $K_p>K_c$

• A. $2{SO}_3\left(g\right)\rightleftharpoons 2{SO}_2\left(g\right)+O_2\left(g\right)$
• B. ${PCl}_3\left(g\right)+{Cl}_2\rightleftharpoons {PCl}_5\left(g\right)$
• C. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• D. $N_2\left(g\right)+{3H}_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

Answer 1

The correct option is A $2{SO}_3\left(g\right)\rightleftharpoons 2{SO}_2\left(g\right)+O_2\left(g\right)$

$K_p=K_c(RT{)}^{\Delta ng}$

For $K_p>K_c\Delta n_g>1$

For (A) $:-{\Delta n}_g=(2+1)-2=1=>K_p>K_c$

For (B) $:-{\Delta n}_g=1-(1+1)=-1=>K_p<K_c$

For (C) $:-{\Delta n}_g=2-(1+1)=-1=>K_p=K_c$

For (D) $:-{\Delta n}_g=2-(3+1)=-2=>K_p<K_c$

Therefore, (A) is the correct answer.