Question

For the reaction, \(2\mathrm{N_2O_5} \rightarrow 4\mathrm{NO_2} + \mathrm{O_2}\), the rate equation can be expressed in two ways \(-\frac{d[\mathrm{N_2O_5}]}{dt} = k[\mathrm{N_2O_5}]^n\) and \(+\frac{d[\mathrm{NO_2}]}{dt} = k'[\mathrm{N_2O_5}]^n\). \(k\) and \(k'\) are related as:

• A. \(k = k'\)
• B. \(2k = k'\)
• C. \(k = 2k'\)
• D. \(k = 4k'\)

Hint:

Express both rates through the unique rate of reaction: -(1/2)d[N2O5]/dt = (1/4)d[NO2]/dt.

Answer 1

The Correct Option is B

Rate of reaction = \(-\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt}\).
So \(\frac{d[\mathrm{NO_2}]}{dt} = 2\left(-\frac{d[\mathrm{N_2O_5}]}{dt}\right)\).
Substituting the given rate laws: \(k'[\mathrm{N_2O_5}]^n = 2k[\mathrm{N_2O_5}]^n\), hence \(k' = 2k\).
Answer: (2) \(2k = k'\)