Question 17.p.8: Calculate the initial concentrations of each sub stance. For......

Calculate the initial concentrations of each sub stance. For part (a) , calculate Q_\text{c}\text{ and compare to given }K_\text{c}.\text{ If }Q_\text{c} > K_\text{c} then the reaction proceeds to the left to make reactants from products. If Q_\text{c} < K_\text{c} then the reaction proceeds to right to make products from reactants. For part (b) , use the result of part (a) and the given equilibrium concentration of \text{PCl}_5 to find the equilibrium concentrations of \text{PCl}_3\text{ and Cl}_2.

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Initial concentrations: \text{[PCl}_5] = 0.1050\text{ mol /0.5000 L} = 0.2100 \ M \\ \text{[PCl}_3] = \text{[Cl}_2 ] = 0.0450\text{ mol/0.5000 L} = 0.0900 \ M \\ \text{a) }Q_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} =\frac{[0.0900][0.0900]}{[0.2100]} =0.038571=0.0386
Q_\text{c}, 0.0386,\text{ is less than }K_\text{c}, 0.042, so the reaction will proceed to the right to make more products.
b) To reach equilibrium , concentrations will increase for the products, \text{PCl}_3\text{ and Cl}_2, and decrease for the reactant , \text{PCl}_5.

\begin{matrix} &&\quad \text{PCl}_5 & \leftrightarrows & \text{PCl}_3 &+&\text{Cl}_2 \\ \text{Initial} &&\quad\quad 0.2100 \ M && 0.0900 \ M & &0.0900 \ M \\ \text{Change} && -\text{x} && +\text{x} && +\text{x} &\end{matrix} \\ \begin{matrix} \overline{\begin{matrix} \text{Equilibrium} && 0.2100-\text{x} & & &0.0900+\text{x}&&0.0900+\text{x} \end{matrix} } \end{matrix}

\text{[PCl}_5] = 0.2065 \ M = 0.2100 – \text{x; x} = 0.0035 \ M \\ \text{[PCl}_3] = [\text{Cl}_2 ] = 0.0900 + \text{x} = 0.0900 + 0.0035 = 0.0935 \ M
Check: Calculate equilibrium constants using the calculated equilibrium concentrations:

K_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} =\frac{[0.0935][0.0935]}{[0.2065]} =0.042335=0.0423

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