Question

In the equilibrium reactionA +3B <-> 2C + 5Dwe mix the following in a 3 L reactor: 1 mole of A, 2 moles of B, 3 moles of C and 4 moles of D. Reactants and products are all gases. When equilibrium is reached at 280 deg C, the concentration of D is found to be 0.75 Molar. What is the value of the equilibrium constant at that temperature?

Answer 1

	A	B	C	D
Initial number of moles	1	2	3	4
No. of moles reacted	x	x	-	-
No. of moles at equilibrium	1-x	2-x	3+2x	4+5x
Equilibrium Concentration	(1-x)/V	(2-x)/V	(3+2x)/v	(4+5x)/V

Given the equilibrium concentration of D is 0.75 Molar. So (5+4x)/V = 0.75, where V is the volume of reactor (3 L).

4+5x=0.75*3=2.25

5x=2.25-4=-1.75

X=(-1.75)/5. The value of x is fund to be -0.35

Substitute the value of x to get the equilibrium concentration.

[A]=0.45, [B]=0.783, [C]=0.767, [D]=0.75

K_c =( [C] ² . [D] ⁵) / ( [A] ¹ . [B] ³)

Substitute the values, K_c = (0.767 ². 0.75 ⁵) / (0.45 ¹ . 0.783 ³)

K_c = 0.646

K_p = K_c * (RT)^∆n, R is the Universal Gas Constant (8.314 J/mol K), T is the temperature in K (280+273=553K),

∆n = Moles of Product – Moles of Reactants = (2+5) - (1+3) = 3

So, K_p = 0.646 * (8.314*553)³

K_p = 6.278 x 10¹⁰