 # Question & Answer: Consider the following reaction at equilibrium: 2H2(g)+S2(g)⇌2H2S(g)+heat In a 10.0-L…..

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?

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Question & Answer: Consider the following reaction at equilibrium: 2H2(g)+S2(g)⇌2H2S(g)+heat In a 10.0-L…..
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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