For the adsorption of hydrogen on platinum, the activation energy is 30 k J mol⁻¹ and for the adsorption of hydrogen on nickel, the activation energy is 41.4 k J mol⁻¹. The logarithm of the ratio of the rates of chemisorption on equal areas of the metals at 300 K is __________ (Nearest integer)
Given: ln 10=2.3
R=8.3 J K⁻¹ mol⁻¹

Solution

Chemistry ma’am · 4 steps

Same metal area means the same A, so the rate ratio depends only on the gap in activation energies.

1. Divide the rates. Write r = A e^(-Eₐ/RT) for each metal. On equal areas the factor A is the same, so it cancels when you divide the platinum rate by the nickel rate.

   log r_(Pt)/r_(Ni) = (E_(a,Ni) - E_(a,Pt))/(2.3 RT)

2. Put in numbers. The gap is 41.4 - 30 = 11.4 kJ mol⁻¹ = 11400 J mol⁻¹. The denominator is 2.3 × 8.3 × 300 = 5727 J mol⁻¹.
3. Divide. 11400/5727 = 1.99. Platinum, with the lower barrier, adsorbs hydrogen about 10² times faster than nickel.
4. So. The logarithm of the ratio of the rates is 1.99, which rounds to 2.

The common slip: 5. Stopping at the natural log gives 11400/(8.3 × 300) = 4.58, which rounds to 5. The question wants the base-10 log, so divide by 2.3 as well.

The idea to keep. When two processes share the same A and T, the rate ratio is e^(Δ Eₐ/RT). Use ln 10 to turn it into a base-10 log, and keep Eₐ in joules.

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