Question 17.80: a) Equilibrium partial pressures for the reactants, nitrogen......
a) Equilibrium partial pressures for the reactants, nitrogen and oxygen, can be assumed to equal their initial partial pressures because the equilibrium constant is so small that very little nitrogen and oxygen will react to form nitrogen monoxide. After calculating the equilibrium partial pressure of nitrogen monoxide, test this assumption by comparing the partial pressure of nitrogen monoxide with that of nitrogen and oxygen.
Based on the small amount of nitrogen monoxide formed, the assumption that the partial pressures of nitrogen and oxygen change to an insignificant degree holds.
\text{P}_\text{nitrogen} \text{(equilibrium)} = (0.780 – 1.33466 \times 10^{-16})\text{ atm = 0.780 atm N}_2 \\ \text{P}_\text{oxygen}\text{ (equilibrium)} = (0.210 – 1.33466 \times 10^{-16})\text{ atm = 0.210 atm O}_2 \\ \text{P}_\text{NO}\text{ (equilibrium) }= 2 ( 1.33466\times 10^{-16})\text{ atm} = 2.66932 \times 10^{-16} = 2.67 \times 10^{-16}\text{ atm NO}
b) The total pressure is the sum of the three partial pressures:
0.780\text{ atm + 0.210 atm} + 2.67 \times 10^{-16}\text{ atm = 0.990 atm} \\ \text{c) }K_\text{p} = K_\text{c}(\text{RT})^{Δn}\quad \text{Δn = 2 mol NO product – 2 mol reactant (1 N}_2 + 1 \text{ O}_2) = 0 \\ K_\text{p} = K_\text{c}(\text{RT})^0 \\ K_\text{c} = K_\text{p} = 4.35 \times 10^{-31} because there is no net increase or decrease in the number of moles of gas in the course of the reaction.