(b) The rate of decrease in N2H4 partial pressure in a closed reaction vessel from the reaction N2H4(g) + H2(g) → 2 NH3(g) is 74 torr per hour. What are the rates of change of NH3 partial pressure and total pressure in the vessel?

Verified step by step guidance

1. Identify the stoichiometry of the reaction: N2H4(g) + H2(g) → 2 NH3(g). This tells us that 1 mole of N2H4 reacts with 1 mole of H2 to produce 2 moles of NH3.
2. Recognize that the rate of decrease in N2H4 partial pressure is given as 74 torr per hour. According to the stoichiometry, for every mole of N2H4 that reacts, 2 moles of NH3 are produced.
3. Calculate the rate of increase in NH3 partial pressure. Since 2 moles of NH3 are produced for every mole of N2H4 that reacts, the rate of increase in NH3 partial pressure is twice the rate of decrease in N2H4 partial pressure.
4. Determine the rate of change of total pressure in the vessel. The total pressure change is the sum of the changes in partial pressures of all gases involved. Consider the stoichiometry and the fact that H2 is consumed and NH3 is produced.
5. Apply the concept of partial pressures and stoichiometry to find the net change in total pressure. Since 1 mole of N2H4 and 1 mole of H2 produce 2 moles of NH3, calculate the net change in moles of gas and relate it to the change in total pressure.