Question

Question: For the reaction 2 A(g) + 3 B(g) + 2 C(g) + 2 D(9), the equilibrium constant at 298 K is 1.378x10-11. If the initial partial pressures of A, and B are 0.457 bar, and 0.355 bar, respectively, determine the partial pressure of D (in bar) once equilibrium has been reached at 298 K. (2 marks)

Answer 1

Ans :-

ICE table is :

...................2A (g)...............+............3B (g) <--------------------> 2C (g)..................+.......................2D (g)

Initial............0.457 bar.....................0.355 bar...........................0 bar.............................................0 bar

Change..........-2y................................-3y.....................................+2y...............................................+2y

Equilibrium.....(0.457-2y) bar.........(0.355-3y) bar........................2y bar..........................................2y bar

Where, y = Amount dissociated per mole

Expression of Equilibrium constant i.e. Kp (which is equal to the product of the partial pressure of products divided by product of the partial pressure of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

Kp = PC2.PD2 / PA2.PB3n

1.378 x 10-11 = (2y)2.(2y)2 / (0.457-2y)2.(0.355-3y) 3

1.378 x 10-11 = 16y4 / (0.457-2y)2.(0.355-3y) 3

If y is very small than neglect y as compare to 0.457 and 0.355

1.378 x 10-11 = 16y4 / (0.457)2.(0.355) 3

16y4 = 1.288 x 10-13

y4 = 1.288 x 10-13 / 16

y = (8.05 x 10-15)1/4

y = 3.0 x 10-4

Therefore, equilibrium partial pressure of D = 2y bar = 2 x 3.0 x 10-4 = 6.0 x 10-4 bar