Initial concentration of reactant in a first order reaction is $0.08 \mathrm{~mol} \mathrm{dm}^{-3}$ What concentration would remain after 40 minute? $\left(\operatorname{given} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=5.00\right)$

Question

Aicharya · Accepted Answer

For a first order reaction, $\begin{aligned} \mathrm{t} & =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\ \therefore \quad \mathrm{k} & =\frac{2.303}{40 \text { minute }} \log _{10} 5=0.04 \end{aligned}$
Now, $\begin{array}{ll} \therefore & 0.04=\frac{2.303}{40} \log _{10} \frac{0.08}{[\mathrm{~A}]_{\mathrm{t}}} \\ \therefore & 0.69=\log _{10}(0.08)-\log _{10}[\mathrm{~A}]_{\mathrm{t}} \\ \therefore & \log _{10}[\mathrm{~A}]_{\mathrm{t}}=-1.096-0.69 \\ \therefore & \log _{10}[\mathrm{~A}]_{\mathrm{t}}=-1.78 \\ \therefore & {[\mathrm{~A}]_{\mathrm{t}}=\operatorname{Antilog}_{10}(-1.78)=0.016} \end{array}$ Alternate method: $\begin{aligned} & \text { Given } \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=5 \\ \therefore \quad & {[\mathrm{~A}]_{\mathrm{t}}=\frac{0.08}{5}=0.016 } \end{aligned}$

Initial concentration of reactant in a first order reaction is $0.08 \mathrm{~mol} \mathrm{dm}^{-3}$ What…

Solution