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In: Chemistry

​Part A Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will...

​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.

Solutions

Expert Solution

queen_honey_blossom answered 2 years ago
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