If the rate of decrease for the partial pressure of N2H4 in a closed reaction vessel is 76 torr/h, what is the rate of change for the partial pressure of NH3 in the same vessel

The reaction is missing and it's;

N2H4 + H2 - --> 2NH3

It asks for the total pressure too.

Answer:

A) Rate of change for NH3 = 152 torr/h

B) The total pressure in the vessel will remain the same.

Step-by-step explanation:

N2H4 + H2 - --> 2NH3

1 mole of N2H4 yields 2 moles of NH3.

From the question, the rate given for N2H4 is 76 torr/h.

Thus, The rate of change for NH3 will be = 2 x 76 = 152 torr/h

Now, on the reaction side, 1 mole of N2H4 reacts with 1 mole of H2. So we have 2 moles on the left hand side.

While on the product side, 2 moles of NH3 are produced.

So the total pressure will remain the same because for every 2 moles on the reaction side, 2 moles are gotten on the product side.