CS2(g) + 3 Cl2(g) CCl4(g) + S2Cl2(g) At a given temperature the reaction above i
ID: 725721 • Letter: C
Question
CS2(g) + 3 Cl2(g) CCl4(g) + S2Cl2(g) At a given temperature the reaction above is at equilibrium when [CS2] = 0.050 M, [Cl2] = 0.25 M, [CCl4] = 0.15 M, [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? To the right. To the left. No change. Cannot predict unless we know the temperature. Cannot predict unless we know whether the reaction is endothermic or exothermic.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