Explain the reaction above is at equalibrium

Reference no: EM13504510

CS2(g) + 3Cl2(g)<-->CCl4(g) + S2Cl2(g) At a given tempature, the reaction above is at equalibrium 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=0.14M, Cl2=).18 M,CCl4=0.28 M, and S2Cl2=0.20 M?

Reference no: EM13504510

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