# Question 45ea5

Feb 12, 2017

1 hour

#### Explanation:

If the reaction is zero order we can write:

$\textsf{\text{Rate} = k {\left[N {H}_{3}\right]}^{0} = k}$

$\textsf{k = 1.50 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{M/s}}$

This means that $\textsf{1.50 \times {10}^{- 6}}$ moles per litre is being lost every second.

The total loss in concentration $\textsf{= \left(6.40 - 1.00\right) {10}^{- 3} = 5.40 \times {10}^{- 3} \text{ ""M}}$

$\therefore$ Time taken = sf((5.40xx10^-3)/(1.50xx10^-6) sf((M)/("M/s")=3.6xx10^3" "s)#

$\textsf{= 60 \textcolor{w h i t e}{x} \text{min}}$

$\textsf{= 1 \textcolor{w h i t e}{x} \text{hr}}$