Question

5. A sealed vessel was filled with 0.400 mol of H2 and 0.500 mol of I2, and the temperature was raised to 870 K. At equilibrium, it was found that 0.786 mol of HI had formed according to the reaction
H2(g) + I2(g) ⇌ HI(g)

(a) Calculate the equilibrium constant Kp at this temperature

Answer 1

   H₂(g)  + I₂(g)    =    2HI(g)

I   0.400   0.500      0

C -X            -X               +2X

E  (0.400-X)  (0.500-X)    2X

2X=0.789moles

X=0.393

Mole of H₂ at equilibrium= (0.400-0.393) =0.007moles

Moles of I₂ at equilibrium= (0.500-0.393) =0.107moles

Since total moles of product is equal to the total sum in the reactant side, Kc=Kp

Kp= (0.786)²/ (0.007×0.107)

Kp=824.83