Question

Calculate the equilibrium constant Kp for this reaction, given the following information (at 296 K ): 2NO(g)+Br2(g)⇌2NOBr(g)Kc=1.8 2NO(g)⇌N2(g)+O2(g)Kc=2.3×1030

Answer 1

Answer:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$

$K_p= 0.074$

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$

$K_p= 2.3\times 10^{30}$

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:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$

Given: Kc = 1.8

Temperature = 296 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1  

Thus, Kp is:

$K_p= 1.8\times (0.082057\times 296)^{-1}$

$K_p= 0.074$

For the second equilibrium reaction:

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$

Given: Kc = $2.3\times 10^{30}$

Temperature = 296 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2) = 0

Thus, Kp is:

$K_p= 2.3\times 10^{30}\times (0.082057\times 296)^{0}$

$K_p= 2.3\times 10^{30}$