For the reaction P → B, the values of frequency factor A and activation energy E A are 4 × 10¹³ s⁻¹ and 8.3 kJ mol⁻¹ respectively. If the reaction is of first order, the temperature at which the rate constant is 2 × 10⁻⁶ s⁻¹ is _____________ × 10⁻¹ K.
(Given : ln 10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹, log2 = 0.30)

Solution

Chemistry ma’am · 4 steps

Take the natural log of the Arrhenius equation, and T is just Eₐ/R divided by ln(A/k).

1. Rearrange Arrhenius. From k = A e^(-Eₐ/RT), divide by k and take the natural log of both sides. Then solve for T.

   T = Eₐ/(R ln(A/k))

2. Find the log. A/k = (4 × 10¹³)/(2 × 10⁻⁶) = 2 × 10¹⁹. So ln(2 × 10¹⁹) = ln 2 + 19 ln 10 = (2.3 × 0.30) + (19 × 2.3) = 0.69 + 43.7 = 44.39.
3. Put in numbers. With Eₐ in joules, Eₐ/R = 8300/8.3 = 1000 K. So T = 1000/44.39 = 22.53 K.
4. So. T = 22.53 K = 225.3 × 10⁻¹ K, so the blank is 225.

The common slip: 23. Writing the temperature in plain kelvin gives about 23, not 225. The blank is in units of 10⁻¹ K, so multiply 22.53 by 10.

The idea to keep. When A, Eₐ and k are given, ln(A/k) = Eₐ/RT gives T in one line. Use ln 2 = 2.3 log 2 when the question gives you only log values.

Answer from the NTA final answer key. Solution by PrepWiser.