Question

For the reaction: $3A_{(g)}\to 2B_{(g)}$ , the rate of formation of ' $B$ ' at $\text{298}\text{K}$ is represented as $\text{ln}\left(\frac{\text{d}\left[\text{B}\right]}{\text{dt}}\right)\text{=}-\text{4}\text{.606}\text{+}\text{2}\text{ln}\left[\text{A}\right]$ . The order of reaction is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$\ln \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=-4.606+2\ln [\mathrm{A}]$
$2.303\log \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=-4.606+2\times 2.303\log [\mathrm{A}]$
$\log \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\frac{-4.606}{2.303}+2\log [\mathrm{A}]$
$$\begin{gathered} \log \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=-2+2\log [\mathrm{A}] \\ \log \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\log 10^{-2}+\log [\mathrm{A}{]}^2 \end{gathered}$$
$\log \left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\log \left(10^{-2}[\mathrm{A}{]}^2\right)$
$\left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\left(10^{-2}[\mathrm{A}{]}^2\right)$
$\text{ Compare with }\left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\mathrm{k}[\mathrm{A}{]}^{\mathrm{m}}$
$\left(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)=\left(10^{-2}[\mathrm{A}{]}^2\right)$
$\text{ Order }=2$