Question

A vessel of volume 22.4 dm3 contains 4.0 mol H2 and 1.0 mol N2 at 0 initially. All the N2 reacted with sufficient H2 to form NH3. Calculate the partial pressures and the total pressure of the final mixture. Assume the gas constant R is 0.08206 dm3 atm/K mol.

Answer 1

To calculate the partial pressures and total pressure of the final mixture, use the ideal gas law equation and the given number of moles and volume. The partial pressure of H2 is 0.375 atm, the partial pressure of N2 is 0.094 atm, and the total pressure is 0.469 atm.

Step-by-step explanation:

To calculate the partial pressures and total pressure of the final mixture, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. First, let's find the total number of moles of gas in the vessel. We have 4.0 mol of H2 and 1.0 mol of N2, so the total number of moles is 4.0 + 1.0 = 5.0 mol. Next, we can calculate the partial pressures of H2 and N2 using the formula P = nRT/V. The volume of the vessel is given as 22.4 dm3, and the gas constant R is 0.08206 dm3 atm/K mol. Plugging in the values, we have:

PH2 = (4.0 mol)(0.08206 dm3 atm/K mol)(0+273 K)/(22.4 dm3) = 0.375 atm

PN2 = (1.0 mol)(0.08206 dm3 atm/K mol)(0+273 K)/(22.4 dm3) = 0.094 atm

Finally, the total pressure of the final mixture is the sum of the partial pressures, so:

Total pressure = PH2 + PN2 = 0.375 atm + 0.094 atm = 0.469 atm