Question

In the formation of NO and $O_2$$\text{from}N{O}_2\{2N{O}_2\to 2NO+O_2\}$, the rate of production of

• A. NO and $O_2$ are equal.
• B. NO is double the rate of consumption of $N{O}_2$.
• C. NO is twice the rate of production of $O_2$.
• D. $O_2$ is twice the rate of production of NO.

Answer 1

The correct option is C NO is twice the rate of production of $O_2$.

From rate law

$-d\left[N{O}_2\right]/2\ast dt=+d\left[NO\right]/2\ast dt=+d\left[O_2\right]/dt$

For two mole consumption of $N{O}_2$, two moles of NO and one mole of $O_2$ are formed.