Question 17.68: a) You are given a value of Kc but the amounts of reactant a......
a) You are given a value of K_\text{c} but the amounts of reactant and product is given in units of pressure.
Convert K_\text{c}\text{ to }K_\text{p}.
b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the moles equals the molarity. Note the 2: 1:2 mole ratio between \text{SO}_2: \text{O}_2: \text{SO}_3
a) K_\text{p} = K_\text{c}(\text{RT})^{Δn} \quad\quad \text{Δn} = 2 – 3 = -1 (2 mol of product, SO_3 and 3 mol of reactants, 2 SO_2 + \text{O}_2) \\ K_\text{p} = K_\text{c}(\text{RT})^{Δn} = K_\text{c}(\text{RT})^{-1} = (1.7 \times 10^8)[(0.0821\text{ L}\cdot\text{atm/mol}\cdot\text{K) (600. K})]^{-1} = 3.451 \times 10^6\text{ (unrounded)} \\ K_\text{p}=\frac{\text{P}^2_{\text{SO}_3}}{\text{P}^2_{\text{SO}_2}\text{ P}_{\text{O}_2}} =\frac{(300.)^2}{\text{P}^2_{\text{SO}_2}(100.)} =3.451\times 10^6 \\ \text{P}_{\text{SO}_2}=0.016149=0.016\text{ atm}
b)
x = 0.0010, therefore:
[\text{SO}_2] = 0.0040 – 2(0.0010) = 0.0020 M
[\text{O}_2] = 0.0028 -0.0010 = 0.0018 M
[\text{SO}_3] = 2(0.0010) = 0.0020 M
Substitute equilibrium concentrations into the equilibrium expression and solve for K_\text{c}.
The pressure of \text{SO}_2 is estimated using the concentration of \text{SO}_2 and the ideal gas law (although the ideal gas law is not well behaved at high pressures and temperatures).
\text{P}_{\text{SO}_2}=\frac{\text{nRT}}{\text{V}} =\frac{(0.0020\text{ mol})\left\lgroup 0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}} \right\rgroup (1000.\text{ K})}{(1.0\text{ L})} =0.1642=0.16\text{ atm}