Question

Rate Determination and Activation Energy DATA TABLE Trial Temperature (°C) Rate constant, (S-1) 0.00157 24 0.007435 2 16 3 11 0.005563 4 1. Po above, using Temperce ahd. the rate сон dik, as they axis. 2. Determine the activation energy, Ea, by plotting the natural log of k vs. the reciprocal of absolute temperature. You can calculate 1/T (first convert T to K) and in k manually, or use Excel to do it. You will also need to make a graph. 3. Calculate the activation energy, Ea, for the reaction. To do this, first, calculate the best fit line equation for the data in Step 2. Use the slope, m, of the linear fit to calculate the activation 4. A well-known approximation in chemistry states that the rate of a reaction often doubles for every 10°C increase in temperature. Use your data to test this rule. (Note: It is not necessarily equal to 2.00; this is just an approximate value and depends on the activation energy for the reaction.) Choose the data that is closest to 10°C apart and check if the rule holds. 5. Using the rate constant and precise temperature value for the trial that was done at room temperature (-20°C), as well as the Ea value you obtained in Step 3 above, calculate the value of the rate constant at 40°C.

Answer 1

ANSWER:

Part 2

• To get 1/T we need to transform first the temperature from ºC to K

$T_{(K)}=T_{(^{\circ}C)}+273.15$

• Then, the data that we need to plot ln K vs 1/T is

Trial	T (ºC)	T (K)	1/T (K-1)	k (s-1)	ln k
1	24	297.15	3.37x10-3	0.008157	-4.8089
2	20	293.15	3.41x10-3	0.007435	-4.9016
3	16	289.15	3.46x10-3	0.006931	-4.9718
4	11	284.15	3.52x10-3	0.005563	-5.1916

• Now, we can made the plot of ln K vs 1/T and add the trendline equation (best-fit line equation) to plot:

In k vs 1/T -4.75 -4.80 -4.85 -4.90 In K -4.95 -5.00 -5.05 -5.10 y = -2521.5x + 3.6862 R2 = 0.9963 -5.15 -5.20 -5.25 3.32E-03 3.36E-03 3.40E-03 3.44E-03 3.48E-03 3.52E-03 3.56E-03 1/T (K-1)

• To get the best correlation factor (R²) and the best fit line equation we do not use the data of trial 3.

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Part 3

• From the plot of ln k vs 1/T, the trendline equation is

y = -2521.5.0 + 3.6862

or

$ln\, k=-2521.5\left ( \frac{1}{T} \right )+3.6862$

• This equation is similar to

$ln\, k=-\frac{Ea}{R}\left ( \frac{1}{T} \right )+ln\, A$

where R = gas constant = 8.314 J/mol.K and A = pre-exponential factor

• Then, using the slope from plot, we can calculate the activation energy

$m=-\frac{Ea}{R}=-2521.5\, K$

then

$\frac{Ea}{R}=2521.5\, K$

$Ea=2521.5\, K\times 8.314\, \frac{J}{mol.K}=\mathbf{20964\, \frac{J}{mol}=20.96\, \frac{KJ}{mol}}$

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Part 4

• The closest data to 10 ºC is the data at 11ºC. To check the rule we use the data at 11 ºC and 20 ºC (almost 10º C mora then 11ºC):

Trial	T (ºC)	k (s-1)
2	20	0.007435
4	11	0.005563

• Rate constant depends on the temperature, usually, increasing temperature cause that rate constant increases. Double the reaction rate by increasing 10 ºC of temperature (and concentrations of reactants don't change) means that constant rate should double when temperature is increased by 0 ºC.
• Then, if the rule holds, the conatant rate ratio must be 2.00 for trial 2 and 4:

$\frac{k_{20^{\circ}C}}{k_{11^{\circ}C}}=\frac{0.007435}{0.005563}=1.34$

• In this case, the rule does not hold because the ratio is 1.34 (it must be 2)

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Part 5

• The data at 20 ºC is

T (ºC)	T (K)	k (s-1)
20	293.15	0.007435
40	313.15	??

• To calculate the rate constant at 40 ºC we use the following equation:

$ln\left ( \frac{k_{1}}{k_{2}} \right )=-\frac{Ea}{R}\left ( \frac{1}{T_{1}}- \frac{1}{T_{2}}\right )$

where T1 = 40 ºC = 313.15 K and T2 = 20 ºC = 293.15 K

from part 3, the Ea = 20964 J/mol

then

$ln\left ( \frac{k_{1}}{0.007435} \right )=-\frac{20964\frac{J}{mol}}{8.314\frac{J}{mol.K}}\left ( \frac{1}{313.15\, K}- \frac{1}{293.15\, K}\right )$

$ln\left ( \frac{k_{1}}{0.007435} \right )=0.5494$

$\frac{k_{1}}{0.007435} =e^{0.5494}=1.7321$

$k_{1} =1.7321\times 0.007435\, s^{-1}=\mathbf{0.012878\, s^{-1}}$