C. Finding the Activation Energy: 5. a. Using the data from your ti at room temp

Question

C. Finding the Activation Energy: 5. a. Using the data from your ti at room temperature), and the data for trials 6-9 which use the same volumes at different T (shared as specified in lab), make an Excel table of k, In k, T, and T. What units must be used for T? Make Excel do these calculations b. Prepare an Arrhenius plot of In k vs 1/T, and include the correlation R2 and the equation of the linear trend-line on your graph. Should yo“ plot the origin or not? Follow our format guidelines! c. Use the slope of the linear fit to calculate the activation energy E in kJ for the reaction. Show this calculation in your lab notebook (Individual). Be sure to include the correct units D. Calculating the Activation Energy when a catalyst is present Calculate the rate and rate constant k for your catalyst trial as before. Use the Arrhenius equation, the results from Part C, and the rate constant and temperature from your catalyst trial to calculate the activation energy for Reaction in the presence of the copper(ion Show this calculation in your lab notebook (Individual). 6. We will assume the frequency factor A is the same with or without a catalyst in order to do this calculation. Therefore you should be able to solve for E, in the presence of the catalyst, if you know k and E, in the absence of the catalyst from your previous results

Explanation / Answer

From Arhenius Equation lnK= lnKo-1(Ea/R) /T

So a plot of lnK Vs 1/T ( T in K) gives a slope of -Ea/R and intercept lnKo

The slope is -Ea/R= -6394

Ea= 6394*8.314 =53160 Joules/mole=53.160 Kj/mole

for calculating the activation energy in the presence of catalyst, avearage rate consant for 20.2 is taken and thw plot of 1/T and K are repeated and the plot of lnK vs 1/T is repeated and shown below

Ea/R= 6104, Ea= 6104*8.314= 50749 J/moles= 50.749 Kj/mole

0.003524

T K 20.2 0.00328 20.2 0.00294 20.2 0.00289 20.2 0.00268 20.2 0.00583 Average

0.003524

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