Question

Consider the reaction H2(g)+Cl2(g) <--> 2HCl(g) which is at equilibrium when the pressure of H2 is 0.200 atm, the pressure of Cl2 is 0.0085 atm, and the pressure of HCL is 0.7500 atm. If 0.57870 atm H2 and 0.5003 atm Cl2 are placed in a different contanier at the same temperature, calculate the equilibrium pressures of all three gases.

Answer 1

equilibrium pressures of all three gases

For the reaction H2+Cl2-à 2HCl

Equilibrium constant= [HCl]2/ [H2][Cl2]

K= (0.75)²/{0.0085*0.2)=33.08

Initial partial pressures [ H2] =0.5787atm [H2] =0.5003

At Equilibrium [HCL] =2x [H2] =0.5787-x and [H2] =0.5003-x

Where x= drop in partial pressure of H2 to establish equilibrium

(2x)2/ {0.5787-x)*(0.5003-x)= 33.09

4x2/ {0.5787-x)*(0.5003-x)= 33.08, x²/{0.5787-x)*(0.5003-x)= 33.09/4=8.2725

This problem can be solved by assuming some value of x and matching LHS and RHS

Which gives x= 0.3963

At Equilibrium [HCL] =2*0.3963=0.7926 atm [H2] = 0.5787-0.3963=0.1824 atm and [CL2] =0.5003-0.3963=0.104 atm