Chemical Kinetics - Result Question 21
####1. $NO _2$ required for a reaction is produced by the decomposition of $N _2 O _5$ in $CCl _4$ as per the equation,
$$ 2 N _2 O _5(g) \longrightarrow 4 NO _2(g)+O _2(g) $$
The initial concentration of $N _2 O _5$ is $3.00 mol L^{-1}$ and it is $2.75 mol L^{-1}$ after 30 minutes. The rate of formation of $NO _2$ is
(2019 Main, 12 April II)
(a) $4.167 \times 10^{-3} mol L^{-1} min^{-1}$
(b) $1.667 \times 10^{-2} mol L^{-1} min^{-1}$
(c) $8.333 \times 10^{-3} mol L^{-1} min^{-1}$
(d) $2.083 \times 10^{-3} mol L^{-1} min^{-1}$
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Answer:
Correct Answer: 1. (b)
Solution:
$$ \text { Key Idea The rate of a chemical reaction means the speed } $$
with which the reaction takes place.
For
$$ R \longrightarrow P $$
Rate of disappearance of $R$
$$ =\frac{\text { Decrease in conc. of } R}{\text { Time taken }}=-\frac{\Delta[R]}{\Delta t} $$
Rate of appearance of $P$
$$ =\frac{\text { Increase in conc. of } P}{\text { Time taken }}=+\frac{\Delta[P]}{\Delta t} $$
Given, $\left[N _2 O _5\right] _{\text {initial }}=3.00 mol L^{-1}$
After $30 min,\left[N _2 O _5\right]=2.75 mol L^{-1}$
$$ \begin{array}{ll} & 2 N _2 O _5(g) \longrightarrow 4 NO _2(g)+O _2(g) \ t=0 & 3.0 M \ t=30 & 2.75 M \end{array} $$
From the equation, it can be concluded that
$$ \begin{aligned} & \quad \frac{1}{2} \times \frac{-\Delta\left[N _2 O _5\right]}{\Delta t}=\frac{1}{4} \times \frac{\Delta\left[NO _2\right]}{\Delta t} \ & =\frac{-\Delta\left[N _2 O _5\right]}{\Delta t}=\frac{-(2.75-3.00) mol L^{-1}}{30} \Rightarrow \frac{0.25}{30} \ & \text { and } \quad \frac{\Delta\left[NO _2\right]}{\Delta t}=-2 \frac{\Delta\left(N _2 O _5\right)}{\Delta t} \Rightarrow \frac{\Delta\left[NO _2\right]}{\Delta t}=-2 \times \frac{0.25}{30} \ & =-1.667 \times 10^{-2} mol L^{-1} min^{-1} \end{aligned} $$
