At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 .

If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

given,

int. moles     0.2          1.0           ----

final moles   (0.2-x)    (1.0-x)        x

total moles are 1.2-x

therefore, from Kp equation we have-

6.8 x 10^2  =                   [x/(1.2-x) * 1]              

[(0.2-x)/(1.2-x)*1][(1-x)/(1.2-x)*1]

solving the above quadratic, we get- x

now partial pressure of H2 S is  (x/1.2-x)*1

which is the required answer

hope this helps

AJ

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