Question

For the reaction: $3A_{\left(\right. g \left.\right)} \rightarrow 2B_{\left(\right. g ⁡ \left.\right)}$ , 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}] $
$ \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 $
$ \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$