CS₂(g)+ 3Cl₂(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,and [S₂Cl₂] = 0.35 M.What will be the direction of the reaction when the reactants and products have the following concentrations: CS₂ = 0.14 M,Cl₂ = 0.20 M,CCl₄ = 0.28 M,and S₂Cl₂ = 0.28 M?

Question

Quizplus · Accepted Answer

When the reactants and products have the given concentrations, the reaction quotient Qc can be calculated as follows: Qc = ([CS₂][Cl₂]^3)/([CCl₄][S₂Cl₂]) Substituting the given values, we get: Qc = (0.14*0.2^3)/(0.28*0.28) = 0.1125 Since Qc is less than the equilibrium constant Kc (which can be calculated using the given equilibrium concentrations), the reaction proceeds to the left to reach equilibrium. Therefore, the answer is A) to the left.

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

CS₂(g)+ 3Cl₂(g) CCl₄(g)+ S₂Cl₂(g)