Tutored Practice Problem 15.4.1 COUNTS TOWARDS CRADIE Predict and calculate the

Question

Tutored Practice Problem 15.4.1 COUNTS TOWARDS CRADIE Predict and calculate the effect of concentration changes on an equilibrium system Close Probl Some PCls is allowed to dissociate into PCl3 and Cl2 at 523 K. At equilibrium, [PCIs] 0.249 M, and [PCIl [Ch) -0.112 M. Additional PCl3 is added so that [PCI3new 0.212 M and the system is allowed to once again reach equilibrium. PCls(g)PCI(g+Cl2(g) K 5.01x102 at 523 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [PC15] = PCI] Check & Submit Answer Show Approach

Explanation / Answer

The equilibrium reaction is given as

PCl5 (g) <======> PCl3 (g) + Cl2 (g)

K = [PCl3][Cl2]/[[PCl5] = 5.01*10-2 at 523 K.

a) When excess PCl3 is added, [PCl3] increases. However, K, being an equilibrium constant, must remain constant at a particular temperature. To keep K constant, [PCl5] must increase proportionately, i.e, the reverse reaction is facilitated, producing more PCl5.

b) Since the reverse reaction produces PCl5, we can simply write the reverse reaction as

PCl3 (g) + Cl2 (g) <======> PCl5 (g)

K’ = 1/K = 1/(5.01*10-2) = 19.96 at 523 K.

We have [PCl3]new = 0.212 M; set up the ICE chart as below.

PCl3 (g) + Cl2 (g) <=======> PCl5 (g)

initial                                     0.212       0.112                         0.249

change                                      -x             -x                              +x

equilibrium                         (0.212 – x)(0.112 – x)                 (0.249 + x)

Plug in the expression for K’ and obtain

19.96 = (0.249 + x)/(0.212 – x)(0.112 – x)

====> 19.96 = (0.249 + x)/(0.023744 – 0.324x + x2)

====> 0.47393 – 6.46704x + 19.96x2 = 0.249 + x

====> 0.22493 – 7.46704x + 19.96x2 = 0

====> 19.96x2 – 7.46704x + 0.22493 = 0

The possible roots of x are given as

Clearly, x = 0.34106 doesn’t makes sense x is clearly greater than the starting concentration of Cl2, which is 0.112 M. Therefore, the correct value of x is 0.033 M. Thus, the equilibrium values are

[PCl5] = (0.249 + 0.033) M = 0.282 M

[PCl3] = (0.212 – 0.033) M = 0.179 M.

[Cl2] = (0.112 – 0.033) M = 0.079 M (ans).

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