Question

In a reaction, $A\to B+C$, the following data were obtained:
$\begin{array}{cccc}tinseconds& 0& 900& 1800\\ ConcentrationofA& 50.8& 19.7& 7.62\end{array}$
Prove that it is a first order reaction.

Answer 1

For First order reaction, we have

$k=\frac{1}{t}log\frac{\left[A_o\right]}{\left[A_t\right]}$ ${}_{\left[A_t\right]\to Finalcase}^{\left[A_o\right]\to Initialcase}$

For $t_1=0$ to $t_2=900sec\Rightarrow t=t_2-t_1=900second$

$\left[A_o\right]=50.8mol{L}^{-1}$ & $\left[A_t\right]=19.7mol{L}^{-1}$

$k=\frac{1}{900}log\frac{50.8}{19.7}=\frac{1}{900}\times 0.4114=0.00045se{c}^{-1}$

For $t_2=900sec$ to $t_3=1800sec\Rightarrow t=t_3-t_2=900second$

$\left[A_0\right]=19.7mol{L}^{-1}$ & $\left[A_t\right]=7.62mol{L}^{-1}$

$k=\frac{1}{900}log\frac{19.7}{7.62}=\frac{1}{900}\times 0.4125=0.00045se{c}^{-1}$

Since k for both the condition are same (approx)

So, the give reaction is first order reaction.