Answer
(a)
$n_{PCl_5} = 0.478$
$n_{PCl_3} = 0.623$
$n_{Cl_2} = 0.0728$
(b)
$n_{PCl_5} = 0.413$
$n_{PCl_3} = 0.198$
$n_{Cl_2} = 0.198$
Work Step by Step
(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$
$n_{PCl_5} = 0.413$
$n_{PCl_3} = 0.198$
$n_{Cl_2} = 0.198$
