## General Chemistry: Principles and Modern Applications (10th Edition)

(a) $n_{PCl_5} = 0.478$ $n_{PCl_3} = 0.623$ $n_{Cl_2} = 0.0728$ (b) $n_{PCl_5} = 0.413$ $n_{PCl_3} = 0.198$ $n_{Cl_2} = 0.198$
(a) 1. Calculate the concentrations: $$\frac{0.550 \space mol}{2.50 \space L} = 0.220 \space M$$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$ 2. At equilibrium, these are the concentrations of each compound: $[ PCl_5 ] = 0.220 \space M - x$ $[ PCl_3 ] = 0.220 \space M + x$ $[ Cl_2 ] = 0 \space M + x$ $$3.8 \times 10^{-2} = \frac{(0.220 + x)(x)}{(0.220 - x)}$$ 3. Solve for x: x = 0.0291 $[ PCl_5 ] = 0.220 - 0.0291 = 0.191 \space M$ $[ PCl_3 ] = 0.220 + 0.0291 = 0.249 \space M$ $[ Cl_2 ] = 0.0291 \space M$ 4. Multiplying all by the volume (2.5 L): $n_{PCl_5} = 0.478$ $n_{PCl_3} = 0.623$ $n_{Cl_2} = 0.0728$ (b) 1. Find the concentrations: $$\frac{0.610 \space mol}{2.50 \space L} = 0.244 \space M$$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$ 2. At equilibrium, these are the concentrations of each compound: $[ PCl_5 ] = 0.244 \space M - x$ $[ PCl_3 ] = 0 \space M + x$ $[ Cl_2 ] = 0 \space M + x$ $$3.8 \times 10^{-2} = \frac{(x)(x)}{(0.244 - x)}$$ 3. Solve for x: x = 0.0791 $M$ $[ PCl_5 ] = 0.244 - 0.0791 = 0.165 \space M$ $[ PCl_3 ] = 0.0791 \space M$ $[ Cl_2 ] = 0.0791 \space M$ $n_{PCl_5} = 0.413$ $n_{PCl_3} = 0.198$ $n_{Cl_2} = 0.198$