Question

Ammonia and oxygen reacts at higher temperature 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 $1.08\times {10}^{-2}$ mol litre${}^{-1}$ in 3 seconds. The rate of the reaction is:

• A. $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}=9\times {10}^{-4}$ mol litre${}^{-1}$sec${}^{-1}$
• B. $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}=10\times {10}^{-4}$ mol litre${}^{-1}$ sec${}^{-1}$
• C. $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}=13\times {10}^{-4}$ mol litre${}^{-1}$sec${}^{-1}$
• D. $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}=20\times {10}^{-4}$ mol litre${}^{-1}$sec${}^{-1}$

Answer 1

The correct option is A $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}=9\times {10}^{-4}$ mol litre${}^{-1}$sec${}^{-1}$

The expression for the rate of the reaction is $r=\frac{1}{4}\frac{d\left[NO\right]}{dt}$

Substitute values in the above expression,

$r=\frac{1}{4}\times \frac{1.08\times {10}^{-2}mollitr{e}^{-1}}{3sec}=9\times {10}^{-4}mollitr{e}^{-1}se{c}^{-1}$