MHT CET · Chemistry · Chemical Kinetics
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 \(\mathrm{k}=\frac{1}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\)
- C \(\mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}}\)
- D \(\mathrm{k}=\frac{\mathrm{t}}{2.303} \times \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}}\)
Step-by-step Solution
Detailed explanation
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 \(C\)
\(
\begin{aligned}
& -[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}-[\mathrm{A}]_0 \\
& {[\mathrm{~A}]_{\mathrm{t}}=[\mathrm{A}]_0-\mathrm{kt}} \\
& \mathrm{y}=\mathrm{C}-\mathrm{mx} \\
& {[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}} \\
& \mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}}
\end{aligned}
\)

\(
\begin{aligned}
& -\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0 \\
& -\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0=\mathrm{k} \\
& -\int \mathrm{d}[\mathrm{A}]=\int \mathrm{kdt} \\
& -[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}+\mathrm{c} \\
& \text { at } \mathrm{t}=0 \quad \quad[\mathrm{~A}]_{\mathrm{t}}=[\mathrm{A}]_0 \\
& -[\mathrm{A}]_0=\mathrm{k} \times 0+\mathrm{C} \\
& \mathrm{C}=-[\mathrm{A}]_0
\end{aligned}
\)
On substituting the value of \(C\)
\(
\begin{aligned}
& -[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}-[\mathrm{A}]_0 \\
& {[\mathrm{~A}]_{\mathrm{t}}=[\mathrm{A}]_0-\mathrm{kt}} \\
& \mathrm{y}=\mathrm{C}-\mathrm{mx} \\
& {[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}=\mathrm{kt}} \\
& \mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}}
\end{aligned}
\)
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