Question

For the following reaction, the rate of disappearance of ammonia is $3.6{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}}$ . Then what is the rate of formation of water

6}}{{\text{H}}_2}{\text{O}}$$

Answer 1

For the reaction between ammonia and oxygen to form nitric oxide, write an expression for the rate of the reaction. Rearrange the expression to obtain an expression for the rate of formation of water in terms of rate of consumption of ammonia.
${\text{rate = }} - \dfrac{1}{4}\dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right]{\text{ = }}\dfrac{1}{6}\dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right]{\text{ = }} - \dfrac{1}{5}\dfrac{d}{{dt}}\left[ {{{\text{O}}_2}} \right]{\text{ = }}\dfrac{1}{4}\dfrac{d}{{dt}}\left[ {{\text{NO}}} \right]$

Complete Step by step answer: Ammonia reacts with oxygen to form nitric oxide and water. Write the balanced chemical reaction as shown below:
${\text{4N}}{{\text{H}}_3}{\text{ + 5}}{{\text{O}}_2}{\text{ }} \to {\text{ 4NO + 6}}{{\text{H}}_2}{\text{O}}$
Write the expression for the rate of the reaction
${\text{rate = }} - \dfrac{1}{4}\dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right]{\text{ = }}\dfrac{1}{6}\dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right]{\text{ = }} - \dfrac{1}{5}\dfrac{d}{{dt}}\left[ {{{\text{O}}_2}} \right]{\text{ = }}\dfrac{1}{4}\dfrac{d}{{dt}}\left[ {{\text{NO}}} \right]$
In the reaction, ammonia and oxygen are consumed. Hence, you have given negative sign to their rates. On the other hand, nitric oxide and water are formed. Hence, you have given positive signs to their rates.
Rearrange above expression to obtain the rate of formation of water
$\dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \dfrac{6}{4} \times \left[ { - \dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right]} \right]$
$\Rightarrow \dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \dfrac{3}{2} \times \left[ { - \dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right]} \right]$ … …(1)
The rate of disappearance of ammonia is $3.6{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}}$
$- \dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right] = 3.6{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}}$ … …(2)
Substitute equation (2) into equation (1) and solve for the rate of formation of water:

$$\dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \dfrac{3}{2} \times \left[ { - \dfrac{d}{{dt}}\left[ {{\text{N}}{{\text{H}}_3}} \right]} \right] \\\ \Rightarrow \dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \dfrac{3}{2} \times 3.6{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}} \\\ \Rightarrow \dfrac{d}{{dt}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = 5.4{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}} \\\$$

Hence, the rate of formation of water is $5.4{\text{ }} \times {\text{ }}{10^{ - 3}}{\text{mol/s}}$

Note: You can define the rate of the reaction as the ratio of the change in the concentration to the time taken for that change. You can consider two types of rates of reaction, the average rate and instantaneous rate. You can apply the average rate over a period of time whereas you can apply the instantaneous rate for a particular time instant.