In: Chemistry
Part A
Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward. Concentration (M) initial: change: equilibrium: [XY] 0.500 −x 0.500−x net→ ⇌ [X] 0.100 +x 0.100+x + [Y] 0.100 +x 0.100+x The change in concentration, x , is negative for the reactants because they are consumed and positive for the products because they are produced.
Based on a Kc value of 0.160 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?
Express the molar concentrations numerically.
Part B
Calculating equilibrium concentrations when the net reaction proceeds in reverse
Consider mixture C, which will cause the net reaction to proceed in reverse.
Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+x←net⇌[X]0.300−x0.300−x+[Y]0.300−x0.300−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 Kc value of 0.160 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?
Express the molar concentrations numerically.