Question

For the reaction, \( \mathrm{N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g)} \), the value of the rate of disappearance of \( \mathrm{N_2O_5} \) is given as \( 5.15 \times 10^{-3} \, \mathrm{mol\,L^{-1}\,s^{-1}} \). The rate of formation of \( \mathrm{NO_2} \) is

Hint:

Multiply rate of disappearance by \( \frac{\text{coefficient of product}}{\text{coefficient of reactant}} \) to get rate of formation.

Answer 1

Concept: For a reaction \( aA \rightarrow bB + cC \), the rate can be expressed as: \[ \text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} \]

Step 1: Identify stoichiometric coefficients.
\[ N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2 \] Here, coefficient of \( N_2O_5 \) is 1, coefficient of \( NO_2 \) is 2.

Step 2: Apply rate relation.
Rate of disappearance of \( N_2O_5 \): \[ -\frac{d[N_2O_5]}{dt} = 5.15 \times 10^{-3} \, mol\,L^{-1}\,s^{-1} \] Rate of formation of \( NO_2 \): \[ \frac{d[NO_2]}{dt} = 2 \times \left( -\frac{d[N_2O_5]}{dt} \right) \] \[ = 2 \times (5.15 \times 10^{-3}) \]

Step 3: Final value.
\[ \frac{d[NO_2]}{dt} = 1.03 \times 10^{-2} \, mol\,L^{-1}\,s^{-1} \]