MHT CET · Chemistry · Chemical Kinetics
For a zero order reaction, \(\mathrm{A} \longrightarrow\) product, concentration of A decreases from \(1.2 \mathrm{moldm}^{-3}\) to \(0.4 \mathrm{moldm}^{-3}\) in 240 second. What is rate constant of the reaction?
- A \(0.1 \mathrm{~mol} \mathrm{dm}^{-3}\) minute \({ }^{-1}\)
- B \(0.2 \mathrm{~mol} \mathrm{dm}^{-3}\) minute \(^{-1}\)
- C \(0.3 \mathrm{~mol} \mathrm{dm}^{-3}\) minute \({ }^{-1}\)
- D \(0.4 \mathrm{~mol} \mathrm{dm}^{-3}\) minute \(^{-1}\)
Answer & Solution
Correct Answer
(B) \(0.2 \mathrm{~mol} \mathrm{dm}^{-3}\) minute \(^{-1}\)
Step-by-step Solution
Detailed explanation
For a zero order reaction,
\(\begin{aligned} \mathrm{k} & =\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}} \\ & =\frac{(1.2-0.4) \text { moldm }^{-3}}{4 \text { minute }(\because 1 \text { minute }=60 \text { second })} \\ & =0.2 \mathrm{~mol} \mathrm{dm}^{-3} \text { minute }^{-1}\end{aligned}\)
\(\begin{aligned} \mathrm{k} & =\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}} \\ & =\frac{(1.2-0.4) \text { moldm }^{-3}}{4 \text { minute }(\because 1 \text { minute }=60 \text { second })} \\ & =0.2 \mathrm{~mol} \mathrm{dm}^{-3} \text { minute }^{-1}\end{aligned}\)
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