If the rate of decrease for the partial pressure of N2H4 in a closed reaction vessel is 80 torr/h, what is the rate of change for the partial pressure of NH3 in the same vessel?

Answer:Pressure of NH3 increases at the rate of 160torr/hr

Explanation:

According to the equation below

N2H4 (g) + H2 (g) = => 2 NH3 (g)

It shows that 1mol of N2H4 will gives 2mol of NH3

Therefore, if the partial pressure rate of N2H4 is 80torr/hr, the partial pressure of NH3 increases by 2*80=160torr/hr