Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. + net Concentration (M) [XY] [X] + [Y] initial: 0.200 0.300 0.300 change: 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 becaus they are consumed. Part C Based on a K, value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

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Based on a Kc value of 0.170 and the data table given, what are the equilibrium conversations of XY, X, and Y, respectively?

Calculating equilibrium concentrations when the net reaction proceeds in reverse
Consider mixture C, which will cause the net reaction to proceed in reverse.
– net
Concentration (M) [XY
[X]
+ [Y]
initial:
0.200
0.300
0.300
change:
+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.
Part C
Based on a K, value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y,
respectively?
Express the molar concentrations numerically.
Transcribed Image Text:Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. – net Concentration (M) [XY [X] + [Y] initial: 0.200 0.300 0.300 change: +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. Part C Based on a K, value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.
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