Question

The following reaction was carried out at ${44}^{o}C:N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$
The concentration of $N{O}_2$ is $6.0\times {10}^{-3}M$ after $10$ minutes of the start of the reaction. Calculate the rate of production of $N{O}_2$ over the first ten minutes of the reaction.

• A. $6.0\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$
• B. $6.0\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$
• C. $5.0\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$
• D. $3.0\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$

Answer 1

Answer: (A)

$6.0\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}$

Given conc. Of $N{O}_2$ is $6\times {10}^{-3}M$ , time $=10$ min

Rate of production $=\frac{dN{O}_2}{dT}=\frac{6\times {10}^{-3}M}{10min}$

Hence rate of production is $6\times {10}^{-4}$ $mol{L}^{-1}mi{n}^{-1}$