Question

ave u 2. In an experiment conducted at 74°C, the equilibrium concentrations of reactants and products for the equation shown below were [CO] = 1.2 x 102M, [Cl2] = 0.054 M and [COCI:] =0.14 M. COCI, (g) - CO(g) + Cl2 (g) a) What is the equilibrium expression for this reaction? K = [co] [cle] [Coco Ke@ 740 b) Calculate the value of the equilibrium constant, Kc. K. (1.24 16 ²) (0.054) (6.14M) (2= 4.6 103 @ 14C) c) Calculate the value of the equilibrium constant, Kp. R = 0.0 T = 14 +2 Kp = kc (Rr. An Kp = 46,163 (0.08206 • 341) Kp=0.13 An= Prodie

would an expert verify if this is right. thank you

Answer 1

a)

Kc is defined as concentration of product by concentration of reactant with each concentration term raised to power that is equal to its stoichiometric coefficient in balanced equation

So,

Kc = [CO][Cl2]/[COCl2]

b)

Kc = [CO][Cl2]/[COCl2]

= (1.2*10^-2)*(0.054) / (0.14)

= 4.6*10^-3

Answer: 4.6*10^-3

c)

T= 74.0 oC

= (74.0+273) K

= 347 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = 1

use:

Kp= Kc (RT)^Δ n

Kp = 4.6*10^-3*(0.08206*347.0)^(1)

Kp = 0.131

Answer: 0.13