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

Answer:

$K_p= 0.00016$

Explanation:

The relation between Kp and Kc is given below:

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

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:

$K_p= 0.50\times (0.082057\times 673)^{-2}$

$K_p= 0.00016$