# Part A The rate constant for a certain reaction is k = 7.00xx10^-3 s^-1. If the initial reactant concentration was 0.600 M, what will the concentration be after 19.0 minutes?

Jul 15, 2016

$\textsf{2.05 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

The units for $\textsf{k}$ tells me that the reaction is 1st order.

So we get:

$\textsf{A \rightarrow \text{products}}$

$\therefore$$\textsf{- \frac{d \left[A\right]}{\mathrm{dt}} = k {\left[A\right]}^{1}}$

Integrating gives:

$\textsf{\ln {\left[A\right]}_{t} = \ln \left[{A}_{0}\right] - k t}$

Putting in the numbers:

$\textsf{\ln {\left[A\right]}_{t} = \ln \left[0.6\right] - \left(7 \times {10}^{- 3} \times 19.0 \times 60\right)}$

$\textsf{\ln \left[{A}_{t}\right] = - 0.511 - 7.98 = - 8.491}$

From which:

$\textsf{{\left[A\right]}_{t} = 2.05 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

This is 1st order exponential decay. The concentration / time graph looks like this:

You will notice that the time taken for $\textsf{A}$ to fall by 1/2 its initial value is constant.

This is a feature of 1st order processes.