Consider mixture B, which will cause the net reaction to proceed forward. Concen

Question

Consider mixture B, which will cause the net reaction to proceed forward. Concentration (M) initial:change:equilibrium:[XY]0.500x0.500xnet[X]0.100+x0.100+x+ [Y]0.100+x0.100+x The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced. Part B Based on a K value of 0.220 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

I need a deatiled expaination on how to solvde this equation? thanks

Explanation / Answer

KC for the given reaction = [X] [Y]/ [XY] = (0.1+x)*(0.1+x)/ (0.5-x)= 0.22

(0.1+x)2/ (0.5-x)= 0.22, when solved using excel, x=0.17

hence at equilibrium [XY] = 0.5-0.17=0.33 , [X]= [Y]=0.1+0.17=0.27

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