Question

If the rate of decrease for the partial pressure of N2H4N2H4 in a closed reaction vessel is 70 torr/htorr/h , what is the rate of change for the partial pressure of NH3NH3 in the same vessel

Answer 1

Answer:

$r_{NH_3}=140torr/h$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$N_2H_4 (g) + H_2 (g) \rightarrow 2 NH_3$

Thus, in terms of pressures, the rate becomes:

$-r_{N_2H_4}=\frac{1}{2} r_{NH_3}$

Thus, the rate of change for the partial pressure of ammonia turns out:

$r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h$

The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.

Best regards.