Calculating equilibrium concentrations when the net reaction proceeds in reverse

Question

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. rightarrow net Concentration (M) [XY] [X] + [Y] initial 0.200 0.300 0.300 change: +x -x -x equilibrium: 0.200 + x 0.300 - x 0.300 - x The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a K_c value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.

Explanation / Answer

The reaction is X+Y<--->XY

KC= [XY]/[X][Y] = (0.2+x)/ (0.3-x)2= 1/0.14= 7.14, when solved using excel, x= 0.0963

At equilibrium, [XY] =0.2+0.0963= 0.2963, [X] =[Y] =0.3-0.0963= 0.2037

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