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Consider the following reaction at equilibrium: 2H2(g)+S2(g)⇌2H2S(g)+heat In a 10.0-L container, an equilibrium mixture contains 2.15 g of H2, 11.5 g of S2 and 69.7 g of H2S.
If a 5.50-L container has an equilibrium mixture of 0.320 mol of H2 and 2.45 mol of H2S, what is the [S2] if temperature remains constant?
Expert Answer
The reaction is,
2H2(g)+S2(g)⇌2H2S(g)
mole sof H2 = 2.15 g / 18 =0.1194 mol
moles of S2 = 11.5 g / 64 = 0.180 mol
mole sof H2S = 69.7 g / 34 = 2.05 mol
K eq = (2.05 mol /10 L )2 / ( (0.1194/10)2*(0.180/10))
=16376
now,
new volume = 5.50 L
16376 = (2.45/5.5)2 / ( ( 0.320/5.5)2 x (S2) )
[S2] = 0.00358 M
moles of S2 = 0.00358 M x 5.5 L = 0.0196 moles
