Question

In which of the following reactions will Kc= Kp?A) 4 NH3(g) + 3 O2(g) ⇌2 N2(g) + 6 H2O(g)B) 2 SO3(g) + 2 NO(g) ⇌2 SO2(g) + 2 NO2(g)C) 4 N2(g) + 2 O2(g) ⇌4 N2O(g)D) 6 SO2(g) + 3 O2(g) ⇌6 SO3(g)E) None of theabove reactions have Kc= Kp

Answer 1

Answer: $2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)$

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

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

where,

$K_p$ = equilibrium constant in terms of partial pressure

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant

T = temperature

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}$

$K_p=K_c$ when $\Delta n_g$ = 0

a) $4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

$\Delta n_g$ = change in number of moles of gas particles = $8-7=1$

b) $2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)$

$\Delta n_g$ = change in number of moles of gas particles = $4-4=0$

c) $4N_2(g)+2O_2(g)\rightleftharpoons 4N_2O(g)$

$\Delta n_g$ = change in number of moles of gas particles = $4-6=-2$

d) $6SO_2(g)+3O_2(g)\rightleftharpoons 6SO_3(g)$

$\Delta n_g$ = change in number of moles of gas particles = $6-9=-3$

Thus for reaction ,$2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)$ , $K_p=K_c$ as $\Delta n_g$ = 0