Question

When the reaction 2H_2S(g)^f 2H_2(g) + S_2(g) is carried out at 1065 degree C, K_P = 0.012. Starting with pure H_2S at 1065 degree, what must the initial pressure of H_2S be if the equilibrated mixture at this temperature is to contain 0.250 atm of H_2(g)? A) 1.06 atm B) 1.86 atm C) 0.94 atm D) 0.90 atm E) 1.52 atm

Answer 1

let the initial pressure of H2S= P

let x = partial pressure of S2 formed

So at equilibrium, partial pressures : H2S= P-2x, H2= 2x, S2= x

given at equilibrium 2x=0.25, x= 0.125 atm

K=0.012= [S2] [H2]²/ [H2S]² , =0.125*(0.25)²/ (P-0.25)² = 0.007813/ (P-0.25)² =0.012

(P-0.25)²= 0.65

P-0.25 = 0.81

P= 0.81+0.25 =1.06 atm ( A is correct)