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Instrument 03

Arrhenius

k at two or more temperatures in; the activation energy, A and k at a new temperature out.

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Input

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The first line, “T (°C), k (1/s)”, was read as column titles and skipped.

Result

Ea = 50.0 kJ mol−1

Pre-exponential factor A
5.76 × 105 s⁻¹
Slope of ln k vs 1/T
−6,014 K
R2
1.0000
k at 70 °C
0.0141 s⁻¹
Rate rise, 25 → 35 °C
×1.92
0.003000.003100.003200.003300.00340-7.0-6.5-6.0-5.5-5.01/T (1/K)ln k
Arrhenius plot. The slope is −Ea/R; the intercept is ln A.

The working

Intermediate values are carried to six significant figures; results are rounded to three at the end.

  1. Temperatures in kelvin (°C + 273.15), then 1/T and ln k:
    T °CT (K)k1/T (K−1)ln k
    25298.150.0013.35402 × 10−3−6.90776
    35308.150.0019243.24517 × 10−3−6.25335
    45318.150.0035543.14317 × 10−3−5.63968
    55328.150.0063223.04739 × 10−3−5.06372
  2. Least-squares line through the 4 points: slope −6,014.12 K, intercept 13.2636.
  3. Ea = −slope × R = −(−6,014.12) × 8.31446 J mol−1 K−1 = 50,004.2 J mol−1 = 50.0 kJ mol−1.
  4. A = eintercept = e13.2636 = 5.76 × 105 s⁻¹.
  5. k at 343.15 K = A e−Ea/RT: first −Ea/RT = −50,004.2 / (8.31446 × 343.15) = −17.5262, then k = 5.7587 × 105 × e−17.5262 = 0.0141 s⁻¹.
  6. Rate rise from 25 °C to 35 °C: k(308.15 K)/k(298.15 K) = exp[(Ea/R)(1/298.15 − 1/308.15)] = exp(6,014.12 × 1.08844 × 10−4) = e0.654598 = 1.92.

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