Question

In the reaction $2N_{2}O_{5}\overset{\quad\quad}{\rightarrow}4NO_{2} + O_{2} + \frac{d\left\lbrack NO_{2} \right\rbrack}{dt}$at any time t was found to be 2.4 x 10-4 mole L-1 min-1 with rate constant 4.4 x 10-4 min-1.Hence $- \frac{d\left\lbrack N_{2}O_{5} \right\rbrack}{dt}$ at the same time t and the corresponding rate constant of the reaction respectively would be

• A. 1.2 x 10-4 mole L-1 min-1, and 2.2 x 10-4 min-1
• B. 1.2 x 10-4 mole L-1 min-1, and 8.8 x 10-4 min-1
• C. 4.8 x 10-4 mole L-1 min-1, and 2.2 x 10-4 min-1
• D. 2.4 x 10-4 mole L-1 min-1, and 4.4 x 10-4 min-1

Answer 1

Answer: (A)

1.2 x 10-4 mole L-1 min-1, and 2.2 x 10-4 min-1

Answer 2

For 4 moles of NO2 formed, 2 moles of N2O5 consumed in a given time interval, so the rate of consumption of N2O5 must be half of the rate of formation of NO2, Rate constants w.r.t. these two rates will also have the same relationship. So. The choice (1) is the correct answer.