Question

${NO_2}$ required for a reaction is produced by the decomposition of ${N2O5}$ in ${CCl4}$ as per the equation ${ 2N2O5(g) -> 4NO2(g) + O2(g).}$ The initial concentration of ${N2O5}$ is $3.00\, mol$ ${L^{-1}}$ and it is $2.75\, mol$ ${L^{-1}}$ after $30$ minutes. The rate of formation of ${NO2}$ is :

• A. $\ce{2.083 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• B. $\ce{4.167 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• C. $\ce{8.333 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}}$
• D. $\ce{1.667 \times 10^{-2} \; mol \; L^{-1} \; min^{-1}}$

Answer 1

Answer: (D)

$\ce{1.667 \times 10^{-2} \; mol \; L^{-1} \; min^{-1}}$

Answer 2

${ 2N_2 O_5 (g) -> 4NO_2(g) + O_2(g)}$ $3.0 M$ $2.75$ M $\frac{-\Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{0.25}{30}$ $\frac{1}{2} \times\frac{-\Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{1}{4} \times\frac{- \Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{1}{4} \times\frac{\Delta \left[NO_{2}\right]}{\Delta t}$ $\frac{\Delta \left[NO_{2}\right]}{\Delta t} = \frac{0.25}{30} \times2 = 1.66 \times10^{-2} M /\min$