Consider the tabulated data collected for the reaction 2N2O(g) → 2N2(g) + O2(g) at 500 °C. The rate law for this reaction is rate = k[N2O].

Time (s) [N2O] (M)
0 0.2400
61.9 0.1200
123.8 0.0600
185.7 0.0300

What is the value of the rate constant (without its unit) at this temperature?

Question

Consider the tabulated data collected for the reaction 2N2O(g) → 2N2(g) + O2(g) at 500 °C. The rate law for this reaction is rate = k[N2O].

Time (s)	[N2O] (M)
0	0.2400
61.9	0.1200
123.8	0.0600
185.7	0.0300

What is the value of the rate constant (without its unit) at this temperature?

BlackTom AI · Accepted Answer

For a first-order reaction the integrated rate law is ln([A]t) = ln([A]0) − kt, so k = (1/t) ln([A]0/[A]t).

1) Relevant knowledge points and concepts
- The rate law given, rate = k[N2O], identifies the reaction as first-order in [N2O].
- For first-order kinetics, the integrated form ln([A]t) = ln([A]0) − kt applies and t1/2 = ln2 / k.
- The data show successive halving of [N2O], so using the half-life relation is convenient.

2) Step-by-step reasoning
- From the table: [N2O] = 0.2400 M at t = 0 and 0.1200 M at t = 61.9 s. That is a single half-life (concentration halves).
- Use t1/2 = ln2 / k → k = ln2 / t1/2.

3) Calculations with intermediate steps
- ln2 = 0.693147...
- t1/2 = 61.9 s
- k = 0.693147 / 61.9 = 0.011195... s^−1
- Rounding to three significant figures gives k = 1.12 × 10^−2 (units omitted per the question).

(As a check using other data pairs:)
- Using 0.2400 → 0.0600 in 123.8 s: k = (1/123.8) ln(0.2400/0.0600) = 1.386294/123.8 = 0.011192 ≈ 1.12 × 10^−2.
- Using 0.2400 → 0.0300 in 185.7 s: k = (1/185.7) ln(0.2400/0.0300) = 2.079442/185.7 = 0.01119 ≈ 1.12 × 10^−2.
These checks confirm consistent first-order behavior and the computed k.

4) Why this answer is correct
- The derived k follows directly from the first-order integrated law and matches all data intervals; the numerical value 1.12 × 10^−2 is the consistent rate constant.

5) Why the other multiple-choice options are incorrect
- 4.86 × 10^−3, 3.88 × 10^−3, and 1.94 × 10^−3 are substantially smaller than the value computed from ln ratios and do not reproduce the observed half-life of 61.9 s.
- 6.73 × 10^−2 is much larger and would give a far shorter half-life than observed.

6) Conclusion
The calculated rate constant at 500 °C is 1.12 × 10–2, matching the provided correct answer.

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