Asked by jessica
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
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
Answers
Answered by
DrRebel
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
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
Answered by
DrRebel
did this work for you?
Answered by
jane
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
Answered by
jane
oh wait sorry responding to wrong question
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.