Given N2H4 + 7H2O2 → 2HNO3 + 8H2O.

Question

4. Given N2H4 + 7H2O2 → 2HNO3 + 8H2O. If the rate of appearance of HNO3 is 0.026 M·min−1 , what is the rate of disappearance of H2O2? What is the rate of change of H2O2?

5. For the reaction of exercise 4, what is the “rate of reaction” (without reference to any particular reactant or product)?

6. How can you use the answer to exercise 5 to quickly determine the rate of appearance of each product and the rate of disappearance of each reactant?

Answer 1

N₂H₄ + 7H₂O₂ → 2HNO₃ + 8H₂O

When dealing with rates of reaction (kinetics), you can always write the rate for the above reaction as

rate = -d[N2H4]/dt = -1/7d[H2O2]/dt = +1/2d[HNO3]/dt = + 1/8 d[H2O]/dt

4). Looking at the coefficients in the balanced equation, we see a ratio for HNO3 to H2O2 of 2 :7. This tells us that for every 7 moles of H2O2 that disappear, 2 moles of HNO3 appear in the same time period. So, the rate of disappearance of H2O2 is 7/2 the rate of appearance of HNO3. The rate of disappearance of H2O2 would therefore be 0.026 x 7/2 = 0.091 M min^-1. If you want to find the rate of disappearance of H2O2 using the above equation, it would look like this: -1/7 d[H2O2]/dt = 1/2 d[HNO3]/dt = 0.026

1/7(x) = 1/2(0.026) and x = 0.091 M min^-1

5). rate of reaction = 0.013 M min^-1

1/2(0.026) = 0.013

6). Simply multiply it by the coefficient. So disappearance of N₂H₄ = 0.013 M min^-1

Disappearance of H₂O₂ = 7 x 0.013 = 0.091 M min^-1

Appearance of HNO₃ = 2 x 0.013 = 0.026 M min^-1

Appearance of H₂O = 8 x 0.013 = 0.104 M min^-1