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)`

Step-by-step 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