Answer
a. $$K_c = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$
b. $[PCl_5] = 0.44 \space mole/L$, $[Cl_2] = 0.16 \space mole/L$
c. $K_c = 0.058$
d. Increase.
Work Step by Step
a. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$
b.
1. Find the amount of $PCl_5$ at equilibrium.
If 0.16 mole of $PCl_3$ was produced, 0.16 mole of $PCl_5$ was consumed.
$n(PCl_5) = 0.60 \space mole - 0.16 \space mole = 0.44 \space mole$
If 0.16 mole of $PCl_3$ was produced, 0.16 mole of $Cl_2$ was also produced.
$C(PCl_5) = \frac{0.44 \space mole}{1.0 \space L} = 0.44 \space mole/L$
$C(Cl_2) = \frac{0.16 \space mole}{1.0 \space L} = 0.16 \space mole/L$
c. Substitute the values and calculate the constant value:
$$K_c = \frac{( 0.16 )( 0.16 )}{( 0.44 )} = 0.058$$
d. Adding some product will shift the equilibrium in the direction of reactants, increasing the concentration of $PCl_5$.
