Question

For the reaction $4NH_{3} + 5O_{2} \to 4NO + 6H_{2}O$, if the rate of disappearance of $NH_{3}$ is $3.6 \times 10^{-3} mol \,L^{-1} s^{-1}$, what is the rate of formation of $H_{2}O$ ?

• A. $5.4 \times 10^{-3} mol\, L^{-1} s^{-1}$
• B. $3.6 \times 10^{-3} mol\, L^{-1} s^{-1}$
• C. $4\times 10^{-4} mol\, L^{-1} s^{-1}$
• D. $0.6 \times 10^{-4} mol\, L^{-1} s^{-1}$

Answer 1

Answer: (A)

$-\frac{1}{4}\frac{d\left[NH_{3}\right]}{dt}=+\frac{1}{6} \frac{d \left[H_{2}O\right]}{dt}$
$\frac{d\left[H_{2}O\right]}{dt}=\frac{6}{4}\times3.6\times10^{-3}$
$=5.4\times10^{-3} mol \, L^{-1}\, s^{-1}$