A sample of N2O4(g) is placed in an empty cylinder at 25C. After equilibrium is
ID: 629286 • Letter: A
Question
A sample of N2O4(g) is placed in an empty cylinder at 25C. After equilibrium is reached the total pressure is 1.5 atm and 16% (by moles) of the original N2O4(g) has dissociated to NO2(g). (a) Calculate the value of Kp for this dissociation reaction at 25C. (b) If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of N2O4(g) and NO2(g). (c) What percentage (by moles) of the original N2O4(g) is dissociated at the new equilibrium position (total pressure = 1.00 atm)?Explanation / Answer
a) Kp = (partial pressure NO2)^2/(partial pressure N2O4) So you need the partial pressures Partial pressure = (mole ratio)*(total pressure) Total pressure you know, it's 1.5 atm To find the mole ratios: You initially have x moles N2O4. .16x decomposes into .32x moles of NO2. Now your total moles is 1.16x, with .84x being N2O4 and .32x being NO2. Now you find the mole ratios. .32x/1.16x = .27586 for NO2 and .84x/1.16x = .72414 for N2O4 Now, find the partial pressures: Pressure NO2 = .27586*1.5 = .41379 atm Pressure N2O4 = .72414*1.5 = 1.08621 atm Now, plugging into the equation for Kp gives: Kp = (.41379)^2/1.08621 = .16 b) Now that you know Kp, you can write the equation for the new partial pressures: .16 = (x)^2/(1-x) Solving for x gives .33, your partial pressure for NO2 1-x = .67, your partial pressure for N2O4 c) Reverse of part a) You know that your partial pressure for NO2 is .33. Since there are 2 moles of NO2 for each mole of N2O4, you need to divide by 2, which gives .165 Setting .165 into the equation we used in part a) gives: .165 = x/(1+x) x = .20, or 20%