Question

For the reaction:
$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

The value of rate of disappearance of $N_2{O}_5$ was found to be $6.25\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
Then choose the correct option:

• A. The rate of formation of $N{O}_2$ is $1.25\times {10}^{-2}mol{L}^{-1}{s}^{-1}$
• B. The rate of formation of $N{O}_2$ is $3.125\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
• C. The rate of formation of $O_2$ is $3.125\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
• D. Both (a) and (c)

Answer 1

The correct option is D Both (a) and (c)

For a given reaction, if:
$aA+bB\to cC+dD$

Overall rate $r$ can be expressed as:

$r=\frac{-1}{a}\times \frac{d\left[A\right]}{dt}=\frac{-1}{b}\times \frac{d\left[B\right]}{dt}=\frac{+1}{c}\times \frac{d\left[C\right]}{dt}=\frac{+1}{d}\frac{d\left[D\right]}{dt}$
where:

$-\frac{d\left[A\right]}{dt}=\text{Rate of disappearance of A}$

$-\frac{d\left[B\right]}{dt}=\text{Rate of disappearance of B}$

$+\frac{d\left[C\right]}{dt}=\text{Rate of formation of C}$

$+\frac{d\left[D\right]}{dt}=\text{Rate of formation of D}$

For:
$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

$r=-\frac{d\left[N_2{O}_5\right]}{dt}=\frac{+1}{2}\times \frac{d\left[N{O}_2\right]}{dt}=+2\times \frac{d\left[O_2\right]}{dt}$

$\text{Rate of formation of}N{O}_2=2\times \text{Rate of disappearance of}{N}_2{O}_5$
$\text{Rate of formation of}N{O}_2=2\times 6.25\times {10}^{-3}mol{L}^{-1}{s}^{-1}=1.25\times {10}^{-2}mol{L}^{-1}{s}^{-1}$

$\text{Rate of formation of}{O}_2=\frac{1}{2}\times \text{Rate of disappearance of}{N}_2{O}_5$
$\text{Rate of formation of}{O}_2=\frac{1}{2}\times 6.25\times {10}^{-3}mol{L}^{-1}{s}^{-1}=3.125\times {10}^{-3}mol{L}^{-1}{s}^{-1}$