Question

In a certain experiment, the rate constant, k, was measured as a function of the temperature, T.  Two data points were as follows: (A) 25°C, 1.1 x 10–5 min–1, and (B) 225°C, 2.4 x 10-2 min–1. The student who performed the experiment then plotted the logarithmic form of the Arrhenius equation and found the plot to be linear.

What value of Ea did the student obtain experimentally for the reaction?

I tried everything to work out and am not sure. I even graphed them as plots.

Answer: 47 kJ mol–1

Answer 1

Given:
k₁ = 1.1 x 10¯ˢ, T = 25⁰C = 298 K
k₂ = 2.4 x 10¯², T = 225⁰C = 498 K

Arrhenius Equation:
ln(k₂/k₁) = [(∆Eₐ/R)((T₂ - T₁)/(T₁·T₂))]

Solve for ∆Eₐ
∆Eₐ = [R· ln(k₂/k₁)·( T₁·T₂)]/[T₂ - T₁]
∆Eₐ = [(0.008314Kj/mol·K)·ln(2.4 x 10¯² min¯¹/1.1 x 10¯ˢmin¯¹)·(298 K)(498K)]/[498 K – 298 K]
∆Eₐ = 47.4 Kj/mol

Answer 2

did this work for you?

Answer 3

yeah i figured out what was wrong I used that same process the entire time but was converting 275kj wrong i was always out by a factor of 10. haha

Answer 4

oh wait sorry responding to wrong question