Ammonia reacts with oxygen giving nitrogen and water. If the rate of formation of $\mathrm{N}_{2}$ is $0.70…

Ammonia reacts with oxygen giving nitrogen and water. If the rate of formation of $\mathrm{N}_{2}$ is $0.70 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, the rate at which $\mathrm{O}_{2}$ is consumed is
  1. $1.05 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  2. $0.70 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  3. $2.10 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  4. $0.35 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

Solution

The reaction is $4 \mathrm{NH}_{3}+3 \mathrm{O}_{2} ightarrow 2 \mathrm{~N}_{2}+6 \mathrm{H}_{2} \mathrm{O}$
$$
\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_{2}ight]}{\mathrm{d} t}=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{O}_{2}ight]}{\mathrm{d} t} . \quad \text { Hence, }-\frac{\mathrm{d}\left[\mathrm{O}_{2}ight]}{\mathrm{d} t}=\frac{3}{2} \frac{\mathrm{d}\left[\mathrm{N}_{2}ight]}{\mathrm{d} t}=\left(\frac{3}{2}ight)\left(0.7 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}ight)=1.05 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
$$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more CHEMICAL KINETICS questions on Aicharya