Question

Ammonia and oxygen react at high temperature as:

$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$.

If rate of formation of NO is $3.6\times {10}^{-3}$ mol $L^{-1}$ $s^{-1}$ then what will be the rate of formation of water?

Answer 1

$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O$

$Rate=\frac{1}{4}\frac{\left[△N{H}_3\right]}{△t}$ where $△N{H}_3$ is change in concentration of $N{H}_3$.

Also, $Rate=\frac{1}{4}\frac{\left[△NO\right]}{\left[△t\right]}$....(i)

$=\frac{1}{6}\frac{\left[△H_{2}O\right]}{\left[△t\right]}$....(ii)

Now, Rate of formation of $NO=\frac{\left[△NO\right]}{△t}$

$\therefore$ Rate of formation of water $=\frac{△H_{2}O}{△t}$

The relation between these can be obtained from (i) and (ii) as

$\frac{1}{4}\frac{\left[△NO\right]}{△t}=\frac{1}{6}\frac{\left[△H_{2}O\right]}{△t}$

$\Rightarrow \frac{△H_{2}O}{△t}=\frac{6}{4}\frac{\left[△NO\right]}{△t}$

$\therefore \frac{△H_{2}O}{△t}=\frac{6}{4}[3.6\times {10}^{-3}]mol{L}^{-1}{s}^{-1}$

$\therefore \frac{△H_{2}O}{△t}=5.4\times {10}^{-3}mol{L}^{-1}{s}^{-1}$

Hence, the rate of formation of water is $5.4\times {10}^{-3}mol{L}^{-1}{s}^{-1}$