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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 :

Answer: [tex]2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)[/tex]

Explanation:

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure

[tex]K_c[/tex] = equilibrium constant in terms of concentration

R = Gas constant

T = temperature

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}[/tex]

[tex]K_p=K_c[/tex] when [tex]\Delta n_g[/tex] = 0

a) [tex]4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]8-7=1[/tex]

b) [tex]2SO_3(g)+2NO(g)\rightleftharpoons 2SO_2(g)+2NO_2(g)[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]4-4=0[/tex]

c) [tex]4N_2(g)+2O_2(g)\rightleftharpoons 4N_2O(g)[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]4-6=-2[/tex]

d) [tex]6SO_2(g)+3O_2(g)\rightleftharpoons 6SO_3(g)[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]6-9=-3[/tex]

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

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