Question

For the reaction, $ \, \, \, \, \, N_2 O_{5 (g)} \longrightarrow 2NO_{2 (g)} +\frac{1}{2} O_{2 (g)}$ The value of rate of disappearance of $N_2 O_5$ is given as 6.25 x $10^{-3} mol \,L^{-1}\, s^{-1}$. The rate of formation of $NO_2$ and $O_2$ is given respectively as

• A. 6.25 x $10^{-3} mol L^{-1} s^{-1} \, \, and \, \, 6.25 \times 10^{-3} mol L^{-1} s^{-1}$
• B. 1.25 x $10^{-2} mol L^{-1} s^{-1} \, \, and \, \, 3.125 \times 10^{-3} mol L^{-1} s^{-1}$
• C. 6.25 x $10^{-3} mol L^{-1} s^{-1} \, \, and \, 3.125 \times 10^{-3} mol L^{-1} s^{-1}$
• D. 1.25 x $10^{-2} mol L^{-1} s^{-1} \, \, and \, \, 6.25 \times 10^{-3} mol L^{-1} s^{-1}$

Answer 1

Answer: (B)

Key Idea Rate of disappearance of reactant = rate of appearance of product

or $-\frac{1}{\text{Stoichiometric coefficient of reactant}} \frac{d [\text{reactant}]}{dt}$

$ \, =+ \frac{1}{\text{Stoichiometric coefficient of product}} \frac{d [\text{product}]}{dt}$

For the reaction,

$ N_2 O_5 (g) \longrightarrow 2NO_2 (g) + \frac{1}{2} O_2 (g)$

$ \frac{-d [N_2 O_5]}{dt}=+\frac{1}{2} \frac{d [NO_2]}{dt}$

$\therefore \frac{d [NO_2]}{dt} =-2 \frac{d [N_2 O_5]}{dt}$

$ =2 \times 6.25 \times 10^{-3} mol \, L^{-1} \, s^{-1}$

$ =12.5 \times 10^{-3} mol \, L{-1} \, s^{-1}$

$ =1.25 \times 10^{-2} mol \, L{-1} \, s^{-1}$

$ \frac{d [O_2]}{dt} =- \frac{d [N_2 O_5]}{dt} \times \frac{1}{2}$

$ =\frac{6.25 \times 10^{-3} mol \, L^{-1} \, s^{-1}}{2}$

$ =3.125 \times 10^{-3} mol \, L^{-1} \, s^{-1}$