Question

A mixture of 0.47 mol of H2 and 3.59 mol of Cl2 in a one liter flask is heated to 2800C. Calculate the equilibrium partial pressures of H2, Cl2, and HCl if total pressure is 2.00 atmos. Kp=193 at 2800C

Answer 1

The reaction is

H2+Cl2----> 2HCl

Equilibrium constant Kp= [PHCl]²/ [PH2] [PCl2]

PHCl, PH2 and PCl2 denote the partial pressures of HCl, H2 and Cl2 respectively.

let x= moles of H2 decomposed

At equilibrium

[H2] = 0.47-x [Cl2] = 3.59-x    and [HCl]= 2x since these are present in 1 liter flask these are also concentrations.

total moles =0.47-x+3.59-x+2x= 4.06

Mole fractions= H2=(0.47-x)/4.06,   Cl2= (3.59-x)/4.06 [HCL]= 2x/(4.06)

Partial pressures= [H2] = 2*(0.47-x)/4.06,[CL2]= 2*(3.59-x)/4.06 [HCl]= 2*2x/(4.06)

Kp= 4x2/ (0.47-x)*3.59-x)= 183

Solving ths by trial and error using solver gives x= 0.468464

Partial pressures= 2*(0.47-0.46464)/4.06=0.000757 atm , [CL2] =2*(3.59-0.468464)/4.06=1.537

Rest is [HCL= 2-1.537-0.000757=0.4615 atm