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 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 at this temperature?

BlackTom AI · Accepted Answer

First, I restate the problem to ensure clarity: the reaction follows a rate law of rate = k[N2O], and from the data, [N2O] halves after each 61.9 s interval (0.2400 → 0.1200 → 0.0600 → 0.0300). This pattern indicates first-order decay with respect to N2O because a constant half-life across these successive intervals is a hallmark of first-order kinetics.

Option 1: 6.73 × 10–2 s–1. To validate a value, we can use the relation for first-order kinetics: t1/2 = ln(2)/k. If t1/2 were 61.9 s, then k = 0.693 / 61.9 ≈ 0.0112 s–1, i.e., 1.12 × 10–2 s–1, not 6.73 × 10–2 s–1. This option overestimates k by roughly a factor of 6, so it is incorrect.

Option 2: 4.86 × 10–3 s–1. Using k ≈ 0.00486 s–1 would give a half-life t1/2 ≈ 0.693 / 0.00486 ≈ 142.5 s, which is not consistent with the observed 61.9 s half-life. This shows the value does not match the data and is incorrect.

Option 3: 1.94 × 10–3 s–1. If k ≈ 0.00194 s–1, then t1/2 ≈ 0.693 / 0.00194 ≈ 357 s, again conflicting with the experimental halving period. Therefore this option is not supported by the data.

Option 4: 3.88 × 10–3 s–1. With k ≈ 0.00388 s–1, t1/2 ≈ 0.693 / 0.00388 ≈ 178 s, which does not align with the observed 61.9 s half-life. This makes the option incorrect.

Option 5: 1.12 × 10–2 s–1. This value yields t1/2 = 0.693 / 0.0112 ≈ 61.9 s, exactly matching the time interval over which the concentration halves in the data. It is fully consistent with the observed first-order decay pattern, so this option is correct.

A quick consistency check from the data can be done by noting that the concentration halves every 61.9 s (0.2400 → 0.1200 → 0.0600 → 0.0300), which is a classic signature of first-order kinetics where [N2O] = [N2O]0 e^(−kt). From this, k = ln(2)/61.9 s ≈ 0.693/61.9 ≈ 0.0112 s–1, i.e., 1.12 × 10–2 s–1, reinforcing that Option 5 is the correct choice.

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 at this temperature?Single choice

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