# Question #a5353

Dec 6, 2015

$\left(a\right)$

$5.4 \times {10}^{- 3}$

$\left(b\right)$

$101 \text{s}$

#### Explanation:

(a)

This is a 1st order reaction so:

$R = k {\left[{N}_{2} {O}_{5}\right]}^{1}$

We can integrate this rate law (I won't go into the maths here) to give:

${N}_{t} = {N}_{0} {e}^{- k t}$

${N}_{t}$ is the no. moles after time $t$.

${N}_{0}$ is the initial no. of moles.

Taking natural logs of both sides$\Rightarrow$

$\ln {N}_{t} = \ln {N}_{0} - k t$

Now we put in the values:

$\ln {N}_{t} = \ln \left(0.015\right) - \left(6.82 \times {10}^{- 3} \times 2.5 \times 60\right)$

(note we must convert "min" to "s" so x t by 60)

$\therefore \ln {N}_{t} = - 5.223$

$\therefore {N}_{t} = 0.0054 \text{s}$

(b)

To get the 1/2 life we use:

${t}_{\frac{1}{2}} = \frac{0.693}{k}$

Again, I won't go into the derivation of this.

$\therefore {t}_{\frac{1}{2}} = \frac{0.693}{6.82 \times {10}^{- 3}} = 101 \text{s}$