Which of the following equations represents integrated rate law for zero order reaction?

• A

  $\mathrm{k}=\frac{[\mathrm{A}]_{\mathrm{t}}-[\mathrm{A}]_0}{\mathrm{t}}$
• B

  $\mathrm{k}=\frac{1}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$
• Correct answer
  C

  $\mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}}$
• D

  $\mathrm{k}=\frac{\mathrm{t}}{2.303} \times \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$

Solution

Let us consider a zero order reaction,
$$ \begin{aligned} & -\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0 \\ & -\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0=\mathrm{k} \\ & -\int \mathrm{d}[\mathrm{A}]=\int \mathrm{kdt} \\ & -[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}+\mathrm{c} \\ & \text { at } \mathrm{t}=0 \quad \quad[\mathrm{~A}]_{\mathrm{t}}=[\mathrm{A}]_0 \\ & -[\mathrm{A}]_0=\mathrm{k} \times 0+\mathrm{C} \\ & \mathrm{C}=-[\mathrm{A}]_0 \end{aligned} $$ On substituting the value of $C$ $$ \begin{aligned} & -[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}-[\mathrm{A}]_0 \\ & {[\mathrm{~A}]_{\mathrm{t}}=[\mathrm{A}]_0-\mathrm{kt}} \\ & \mathrm{y}=\mathrm{C}-\mathrm{mx} \\ & {[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}} \\ & \mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}} \end{aligned} $$

Hence, the correct answer is option: 'C'.