At a given temperature the reaction CS2(g) + 3Cl2(g) = CCl4(g) + S2Cl2(g) is at

Question

At a given temperature the reaction
CS2(g) + 3Cl2(g) = CCl4(g) + S2Cl2(g)
is at equilibrium when [CS2] = 0.050 M, [Cl2] = 0.25 M, [CCl4] = 0.15 M, and [S2Cl2] = 0.35 M. What would be the direction of the reaction when the reactants and products have the following concentrations: CS2 = 0.15 M, Cl2 = 0.20 M, CCl4 = 0.30 M, and S2Cl2 = 0.28 M?
A to the right
B no change
C cannot predict unless we know the temperature
D cannot predict unless we know whether the reaction is endothermic or exothermic
E to the left

Explanation / Answer

CS2(g) + 3Cl2(g) = CCl4(g) + S2Cl2(g)

Keq = [CCl4][S2Cl2]/[CS2][Cl2]3

Keq = (0.15 * 0.35) / (0.05 * 0.25 * 0.25 * 0.25)

Keq = 67.2

also at any time of the reaction we can find Qvalue by

Q value = [CCl4][S2Cl2]/[CS2][Cl2]3

= (0.3 * 0.28) / (0.15 * 0.2 * 0.2 * 0.2)

Q value = 70

as Q value is greater then Keq value therefore the reaction will shift in backward direction.

therefore option E is correct. reaction will go towards LEFT

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