Question

If the rate of disappearance of \(\mathrm{N_2O_5}\) in the following reaction is \(1.2\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}}\), the rate of production of \(\mathrm{NO_2}\) in \(\mathrm{mol\ L^{-1}\ s^{-1}}\) is
\[ 2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g) \]

• A. \(1.2\times10^{-5}\)
• B. \(3.6\times10^{-5}\)
• C. \(2.4\times10^{-5}\)
• D. \(4.8\times10^{-5}\)

Hint:

For a balanced reaction, rates of disappearance and formation are related by stoichiometric coefficients. If \[ aA\rightarrow bB, \] then \[ \frac{d[B]}{dt}=\frac{b}{a}\left(-\frac{d[A]}{dt}\right). \]

Answer 1

The Correct Option is C

Step 1: Write the stoichiometric rate relationship.
For 2N\(_2\)O\(_5\) \( o\) 4NO\(_2\) + O\(_2\): rate = \(-rac{1}{2}rac{d[N_2O_5]}{dt} = +rac{1}{4}rac{d[NO_2]}{dt}\).

Step 2: Use the ratio of coefficients.
Rate of production of NO\(_2\) = \(rac{4}{2}\) \( imes\) rate of disappearance of N\(_2\)O\(_5\) = \(2 imes 1.2 imes10^{-5} = 2.4 imes10^{-5}\) mol L\(^{-1}\) s\(^{-1}\). \[ oxed{2.4 imes10^{-5}} \]