To determine the rate of production of [tex]\(NO_2\)[/tex] and [tex]\(O_2\)[/tex] from the decomposition of [tex]\(N_2O_5\)[/tex], we can use stoichiometry based on the given chemical equation:
[tex]\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \][/tex]
Given:
- The decomposition rate of [tex]\(N_2O_5\)[/tex] is [tex]\(0.1 \, M/s\)[/tex].
- Note that the rate is negative because the reactant [tex]\(N_2O_5\)[/tex] is being consumed.
Step 1: Determine the mole ratio from the balanced chemical equation.
- From the equation [tex]\(2 N_2O_5 \rightarrow 4 NO_2 + O_2\)[/tex], we see:
- For every 2 moles of [tex]\(N_2O_5\)[/tex] decomposed, 4 moles of [tex]\(NO_2\)[/tex] are produced.
- For every 2 moles of [tex]\(N_2O_5\)[/tex] decomposed, 1 mole of [tex]\(O_2\)[/tex] is produced.
Step 2: Calculate the rate of production of [tex]\(NO_2\)[/tex].
- The formation rate of [tex]\(NO_2\)[/tex] is twice the rate of decomposition of [tex]\(N_2O_5\)[/tex], because 4 moles of [tex]\(NO_2\)[/tex] are produced for every 2 moles of [tex]\(N_2O_5\)[/tex] consumed.
- Therefore, the rate of production of [tex]\(NO_2\)[/tex] is:
[tex]\[
\text{rate}_{NO_2} = 0.1 \, M/s \times \frac{4}{2} = 0.2 \, M/s
\][/tex]
Step 3: Calculate the rate of production of [tex]\(O_2\)[/tex].
- The formation rate of [tex]\(O_2\)[/tex] is half the rate of decomposition of [tex]\(N_2O_5\)[/tex], because 1 mole of [tex]\(O_2\)[/tex] is produced for every 2 moles of [tex]\(N_2O_5\)[/tex] consumed.
- Therefore, the rate of production of [tex]\(O_2\)[/tex] is:
[tex]\[
\text{rate}_{O_2} = 0.1 \, M/s \times \frac{1}{2} = 0.05 \, M/s
\][/tex]
Thus, the rate of production of [tex]\(NO_2\)[/tex] is [tex]\(0.2 \, M/s\)[/tex] and the rate of production of [tex]\(O_2\)[/tex] is [tex]\(0.05 \, M/s\)[/tex].
The correct answer is:
B) [tex]\(NO_2\)[/tex] is produced at [tex]\(0.2 \, M/s\)[/tex], and [tex]\(O_2\)[/tex] is produced at [tex]\(0.05 \, M/s\)[/tex].
