Question

The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g)...

The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K:

H2(g) + I2(g) ---------->2HI(g)

1) Calculate the equilibrium concentrations of reactants and product when 0.309 moles of H2 and 0.309 moles of I2 are introduced into a 1.00 L vessel at 698 K.

[H2] = M?
[I2] = M?
[HI] = M?

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

  • PCl5(g)------->PCl3(g) + Cl2(g)


An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.210 M PCl5, 5.02×10-2 M PCl3 and 5.02×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 4.12×10-2 mol of PCl3(g) is added to the flask?

[PCl5] = M?
[PCl3] = M?
[Cl2] = M?

3.The equilibrium constant, K, for the following reaction is 5.01×10-2 at 538 K:

  • PCl5(g) ------------->PCl3(g) + Cl2(g)


An equilibrium mixture of the three gases in a 16.4 L container at 538 K contains 0.153 M PCl5, 8.74×10-2 M PCl3 and 8.74×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 7.71 L?

[PCl5] = M?
[PCl3] = M?
[Cl2] = M?

4. The equilibrium constant, Kp, for the following reaction is 55.6 at 698 K:

H2(g) + I2(g) ----------->2HI(g)


If ΔH for this reaction is -10.4 kJ, what is the value of Kp at 623 K?

Kp = ?

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Answer #1

1)       H2(g) + I2(g) ----------> 2HI(g)

initial 0.309 M 0.309 M              -

change   -x       -x                 +2x

equil 0.309-x 0.309-x               2x

Kc = [HI]^2/[H2][I2]

55.6 = (2X)^2/(0.309-X)^2

at equilibrium,

[HI] = 2X = 2*0.244 = 0.488 M

[I2] = 0.309-0.244 = 0.065 M

[H2] = 0.309-0.244 = 0.065 M


2)

            
           PCl5(g)-------> PCl3(g) + Cl2(g)

equil   0.21 M           0.0502 M   0.0502 M

added      -             0.0412 M     -

change   +x               -x           -x

equil   0.21+x          0.0914-x     0.0502-x

Kc = [PCl3][Cl2]/[PCl5]

1.2*10^-2 = (0.0914-x)(0.0502-x)/(0.21+x)

x = 0.015

at equilibrium,

[PCl3] = 0.0914-0.015 = 0.0764 M

[Cl2] = 0.0502-0.015 = 0.0352 M

[Pcl5] = 0.21+0.015 = 0.225 M

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