Question

onvert the value of Kc to a value of Kp for the following reaction: N2(g)+3H2(g)⇌2NH3(g) Kc = 0.50 at 400 °C.

Answer 1

`K_p= 0.00016`

Step-by-step explanation:

The relation between Kp and Kc is given below:

`K_p= K_c* (RT)^(\Delta n)`

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

`N_2_((g))+3H_2_((g))\rightleftharpoons2NH_3_((g))`

Given: Kc = 0.50

Temperature =
`400^oC=[400+273]K=673K`

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

`K_p= 0.50* (0.082057* 673)^(-2)`

`K_p= 0.00016`