The equilibrium constant, K, for the following reaction is 1.20×10 -2 at 500 K.

Question

The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K.

An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.325 M PCl5, 6.24×10-2 M PCl3 and 6.24×10-2 MCl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.59×10-2 mol of PCl3(g) is added to the flask?

The equilibrium constant, K, for the following reaction is 1.80×10-4 at 298 K.

An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 298 K contains 0.359 mol NH4HS,   1.34×10-2 M NH3and 1.34×10-2 M H2S. If the concentration of NH3(g) is suddenly increased to 2.42×10-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished?

Explanation / Answer

Part A) PCl5 <==> PCl3 + Cl2

Let x amount of PCl3 has been converted to PCl5 at equilibrium

then at equilibrium,

[PCl5] = 0.325 + x

[PCl3] = (0.0983 - x)

[Cl2] = 6.24 x 10^-2 - x

So,

K = [PCl3][Cl2]/[PCl5]

1.2 x 10^-2 = (0.0983 - x)(6.24 x 10^-2 - x)/(0.325 + x)

3.9 x 10^-3 + 1.2 x 10^-2x = 6.134 x 10^-3 - 0.1607x + x^2

x^2 - 0.1727x + 2.234 x 10^-3 = 0

x = 0.0141 M

So we have at equilibrium,

[PCl5] = 0.325 + 0.0141 = 0.3391 M

[PCl3] = (0.0983 - 0.0141) = 0.0842 M

[Cl2] = 6.24 x 10^-2 - 0.0141 = 0.0483 M

Part B) For NH4HS <==> NH3 + H2S

K = [NH3][H2S]

Initial

[NH3] = 2.42 x 10^-2 + 3.59 x 10^-2 = 0.0376 M

[H2S] = 1.34 x 10^-2 M

Let x be the change in concentration at equilibrium

1.8 x 10^-4 = (0.0376 - x)(1.34 x 10^-2 - x)

x^2 - 0.051x + 3.24 x 10^-4 = 0

x = 7.44 x 10^-3 M

therefore at equilibrium,

[NH3] = 0.0376 - 7.44 x 10^-3 = 0.03016 M

[H2S] = 1.34 x 10^-2 - 7.44 x 10^-3 = 5.96 x 10^-3 M

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