In the correct plot from the question above, the slope of the curve is determine
ID: 811692 • Letter: I
Question
In the correct plot from the question above, the slope of the curve is determined to be -7540 K^1. Note that natural logarithms were used when required. What is the activation energy for this reaction? -62.7 kJ-mol^-1 -907 J-mol^-1 -144 kJ mol^-1 907 J mol^-1 144 kJ mol^-1 62.7 kJ mol^-1 The reaction is measured at three temperatures and the following rate constants have been determined: Temperature (K) 277 300 319 Rate Constant k (M1 .s1) 277 5.43 x 10^-4 4.00 x 10^-3 2.00 x 10^-2 What would you graph if you needed to extract the activation energy from the slope of the graph? (x-axis vs. y-axis) T vs ln(k) T vs log(k) T vs. k 1/T vs 1/k 1/T vs. In(k)Explanation / Answer
1) 1/T vs lnK
2) slope = -E/R
Thus, activation energy,E = -slope*R = 7540*8.314 = 62687.56 J/mole = 62.7 kJ/mole