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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,
H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2
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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H2(g)+S(s)⇔H2S(g)0.2 2 00.2-x 2-x x ∵Δn=0,Kp=Kc Kc=[H2][H2S]=(0.2−x)x=6.8×10−2 On solving this, we get =>x=1.27×10−2 PH2S=[x]RT=1.27×10−2×0.0821×(273+90)=0.38atm
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