1/T in 1/K In K 0.0035 5.65 0.0034 5.68 0.00337 5.69 0.00331 5.71 0.00321 5.74 y= 312.45x + 6.7432 11 A.) Calculate the Activation Energy (Ea ) for the reaction. Ea should be in kJ/mol. This means you should use R in kJ/mol as well (8.314 x 10-3 kJ/mol). Show all work. B.) Calculate the preexponential factor (A) from the trendline equation. Show all your work. C.) As studied in class, the exponential factor in the Arrhenius Equation, , equals the fraction of collisions that have the needed activation energy. D.) Use the Ea you obtained in this experiment (#11a) to calculate this fraction for the highest temperature run – (approx.) 38.0o Remember that “e to the x” is the inverse of the ln function on some calculators. Also remember that, this being a fraction, you should obtain a number between zero and 1, in this case quite a bit closer to zero! Since Ea is in kJ/mol, use the Gas Constant, R, as 8.314 x 10-3 kJ/mol K. The temperature should be in K. e.) Calculate this fraction at the lowest temperature of the experiments, approx. 11.0o f.) Does the ratio of these fractions reflect the ratio of the k’s that were found? In other words, does the dependence of the fraction of collisions with the needed Ea on temperature completely explain the change in rate constant with temperature? g.) Find the ratio of the rate of reaction with the catalyst to the rate of reaction without the catalyst. Comment on what this ratio tells you about the effect of a catalyst on a reaction rate. h.) Use your result for Ea and the fact that H for this reaction is -350. kJ to sketch a reaction profile diagram for this system. (This is a diagram similar to those given on page 1 of this handout.) Show the effect of the catalyst on the reaction profile.
| 1/T in 1/K | In K |
| 0.0035 | 5.65 |
| 0.0034 | 5.68 |
| 0.00337 | 5.69 |
| 0.00331 | 5.71 |
| 0.00321 | 5.74 |
y= 312.45x + 6.7432
11 A.) Calculate the Activation Energy (Ea ) for the reaction. Ea should be in kJ/mol. This means you should use R in kJ/mol as well (8.314 x 10-3 kJ/mol). Show all work.
B.) Calculate the preexponential factor (A) from the trendline equation. Show all your work.
C.) As studied in class, the exponential factor in the Arrhenius Equation, , equals the fraction of collisions that have the needed activation energy.
D.) Use the Ea you obtained in this experiment (#11a) to calculate this fraction for the highest temperature run – (approx.) 38.0o Remember that “e to the x” is the inverse of the ln function on some calculators. Also remember that, this being a fraction, you should obtain a number between zero and 1, in this case quite a bit closer to zero! Since Ea is in kJ/mol, use the Gas Constant, R, as 8.314 x 10-3 kJ/mol K. The temperature should be in K.
e.) Calculate this fraction at the lowest temperature of the experiments, approx. 11.0o
f.) Does the ratio of these fractions reflect the ratio of the k’s that were found? In other words, does the dependence of the fraction of collisions with the needed Ea on temperature completely explain the change in rate constant with temperature?
g.) Find the ratio of the rate of reaction with the catalyst to the rate of reaction without the catalyst.
Comment on what this ratio tells you about the effect of a catalyst on a reaction rate.
h.) Use your result for Ea and the fact that H for this reaction is -350. kJ to sketch a reaction profile diagram for this system. (This is a diagram similar to those given on page 1 of this handout.) Show the effect of the catalyst on the reaction profile.
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