Question

2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.066 M/s. a. At what rate is NO2 being formed? (3 points) b. At what rate is molecular oxygen reacting? (3 points)

Answer 1

Answer :

(a) The rate of $NO_2$ formed is, 0.066 M/s

(b) The rate of $O_2$ formed is, 0.033 M/s

Explanation : Given,

$\frac{d[NO]}{dt}$ = 0.066 M/s

The balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $O_2$ = $-\frac{d[O_2]}{dt}$

The rate of formation of $NO_2$ = $\frac{1}{2}\frac{d[NO_2]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ = 0.066 M/s

(a) Now we have to determine the rate of $NO_2$ formed.

$\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s$

The rate of $NO_2$ formed is, 0.066 M/s

(b) Now we have to determine the rate of molecular oxygen reacting.

$-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

The rate of $O_2$ formed is, 0.033 M/s

Answer 2

The rate of the formation is the time taken and required by the reaction to yield the product. The rate of formation of nitrogen dioxide is, 0.066 M/s and oxygen is 0.033 M/s.

What is the rate of formation?

The rate of formation is the time derivative of the chemical reaction. The balanced chemical reaction can be given as:

$\rm 2 NO + O_{2} \rightarrow 2NO_{2}$

From the reaction:

• The rate of disappearance of $\rm NO$ = $-\dfrac{1}{2} \dfrac{d[\rm NO]}{dt}$
• The rate of disappearance of $\rm O_{2}$ =  $-\dfrac{d[\rm O_{2}]}{dt}$
• The rate of formation of $\rm NO_{2}$ = $\dfrac{1}{2} \dfrac{d[\rm NO_{2}]}{dt}$

Given,

Rate of reaction  $\dfrac{d[\rm NO]}{dt}$ = 0.066 M/s

Calculate the rate of formation of $\rm NO_{2}$ :

$\begin{aligned}\dfrac{1}{2} \dfrac{d[\rm NO_{2}]}{dt} &= \dfrac{1}{2} \dfrac{d[\rm NO]}{dt} \\\\\dfrac{d[\rm NO_{2}]}{dt} &=\dfrac{d[\rm NO]}{dt}\\\\&= 0.066 \;\rm M/s\end{aligned}$

Calculate the rate for molecular oxygen:

$\begin{aligned}-\dfrac{d[\rm O_{2}]}{dt} &= -\dfrac{1}{2} \dfrac{d[\rm NO]}{dt} \\\\\dfrac{d[\rm O_{2}]}{dt} &=\dfrac{1}{2} \times 0.066 \\\\&= 0.033\;\rm M/s\end{aligned}$

Therefore, the rate of formation of nitrogen dioxide is, 0.066 M/s and oxygen is 0.033 M/s.

Learn more about the rate of formation here:

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