Chemical Kinetics Question 8
Question 8 - 2024 (30 Jan Shift 2)
$NO _2$ required for a reaction is produced by decomposition of $N _2 O _5$ in $CCl _4$ as by equation
$2 N _2 O _5(g) \rightarrow 4 NO _2(g)+O _2(g)$
The initial concentration of $N _2 O _5$ is $3 mol L^{-1}$ and it is $2.75 mol L^{-1}$ after 30 minutes.
The rate of formation of $NO _2$ is $x \times 10^{-3} mol L^{-1} min^{-1}$, value of $x$ is _________
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Answer (17)
Solution
Rate of reaction (ROR)
$=-\frac{1}{2} \frac{\Delta\left[N _2 O _5\right]}{\Delta t}=\frac{1}{4} \frac{\left[NO _2\right]}{\Delta t}=\frac{\Delta\left[O _2\right]}{\Delta t}$
$ROR=-\frac{1}{2} \frac{\Delta\left[N _2 O _5\right]}{\Delta t}=-\frac{1}{2} \frac{(2.75-3)}{30} mol L^{-1} min^{-1}$
$ROR=-\frac{1}{2} \frac{(-0.25)}{30} molL^{-1} \min ^{-1}$
$ROR=\frac{1}{240} molL^{-1} \min ^{-1}$
Rate of formation of $NO _2=\frac{\Delta\left[NO _2\right]}{\Delta t}=4 \times ROR$
$=\frac{4}{240}=16.66 \times 10^{-3} molL^{-1} min^{-1} \simeq 17 \times 10^{-3}$.
