Question 17.73: a) Calculate the partial pressures of oxygen and carbon diox......
a) Calculate the partial pressures of oxygen and carbon dioxide because volumes are proportional to moles of gas, so volume fraction equals mole fraction. Assume that the amount of carbon monoxide gas is small relative to the other gases, so the total volume of gases equals \text{V}_{\text{CO }_2}+ \text{V}_{\text{O}_2}+\text{V}_{\text{N}_2}= 10.0 + 1.00 + 50.0 = 61.0.
b) Use the partial pressures and given K_\text{p}\text{ to find P}_\text{CO}. \\ \text{2 CO}_2\text{(g)} \leftrightarrows 2\text{ CO(g) + O}_2\text{(g)} \\ K_\text{p}=\frac{\text{P}_\text{CO}^2\text{ P}_{\text{O}_2}}{\text{P}_{\text{CO}_2}^2} =\frac{\text{P}_\text{CO}^2(0.06557377)}{(0.6557377)^2} =1.4\times 10^{-28} \\ \text{P}_\text{CO}=3.0299\times 10^{-14}=3.0\times 10^{-14}\text{ atm}
c) Convert partial pressure to moles per liter using the ideal gas equation, and then convert moles of CO to grams.
\text{n}_\text{CO}/\text{V = P / RT}=\frac{(3.0299\times 10^{-14}\text{ atm})}{\left\lgroup 0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}} \right\rgroup(800\text{ K})} =4.6131\times 10^{-16}\text{ mol/L (unrounded)} \\ \left\lgroup\frac{4.6131\times 10^{-16}\text{ mol CO}}{\text{L}} \right\rgroup \left\lgroup\frac{28.01\text{ g CO}}{1\text{ mol CO}} \right\rgroup \left\lgroup\frac{1\text{ pg}}{10^{-12}\text{ g}} \right\rgroup =0.01292=0.013\text{ pg CO/L}