Question

Ammonia and oxygen reacts at higher temperatures as
$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$
In an experiment, the concentration of NO increases by $2.16\times {10}^{–2}mollitr{e}^{–1}$ in 3 seconds.

Which of the following is correct?

• A. Rate of the reaction is $7.2\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
• B. Rate of disappearance of ammonia is $7.2\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
  
• C. Rate of formation of water is $4.32\times {10}^{-2}mol{L}^{-1}{s}^{-1}$
• D. Rate of disappearance of oxygen is $3.6\times {10}^{-2}mol{L}^{-1}{s}^{-1}$

Answer 1

The correct option is B Rate of disappearance of ammonia is $7.2\times {10}^{-3}mol{L}^{-1}{s}^{-1}$


The rate expression can be written as,
$-\frac{1}{4}\frac{\Delta \left[N{H}_3\right]}{\Delta t}=-\frac{1}{5}\frac{\Delta \left[O_2\right]}{\Delta t}=\frac{1}{4}\frac{\Delta \left[\mathrm{NO}\right]}{\Delta t}=\frac{1}{6}\frac{\Delta \left[H_{2}O\right]}{\Delta t}$
$r_{reaction}=\frac{1}{4}\frac{d\left[NO\right]}{dt}=\frac{2.16\times {10}^{-2}}{4\times 3}=1.8\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
Rate of disappearance of ammonia is = $r_{reaction}\times 4=7.2\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
Rate of formation of water is = $r_{reaction}\times 6=1.08\times {10}^{-2}mol{L}^{-1}{s}^{-1}$
Rate of disappearance of oxygen is = $r_{reaction}\times 5=9.0\times {10}^{-3}mol{L}^{-1}{s}^{-1}$