Question

The equilibrium constant for the chemical equation N2(g)+3H2(g)↽−−⇀2NH3(g) N 2 ( g ) + 3 H 2 ( g ) ↽ − − ⇀ 2 NH 3 ( g ) is Kp = 0.174 0.174 at 243 243 °C. Calculate the value of Kc for the reaction at 243 243 °C.

Answer 1

Answer:

$K_c=312.13$

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:

$N_2_{(g)}+3H_2_{(g)}\rightleftharpoons 2NH_3_{(g)}$

Given: Kp = 0.174

Temperature = 243 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (243 + 273.15) K = 516.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

$0.174= K_c\times (0.082057\times 516.15)^{-2}$

$K_c\frac{1}{1793.83764}=0.174$

$K_c=0.174\times \:1793.83764$

$K_c=312.13$