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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?

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?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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