1/T (K-¹)2 N₂O5(g) → 4NO2(g) + O₂(g)K(S-¹)4.87 x 10-31.50 x 10-³4.98 x 10-41.35 x 10-43.46 x 10-57.87 x 10-7T (K)3380.00303280.00303180.00313080.00322980.00342736.0037The data in the table above provide the temperature dependence of the rate constant for the reactionshown. Complete the table with required information to 2 significant figures each.- a=-12571.42In K, no units5.3- 6.57.68.9Carefully construct a neat and well labelled graph/plot of In K (y-axis) versus 1/T (x-axis) in the appropriatequadrant (s), using EXCEL or the template below; then determine the slope and activation energy, Ea for the givenreaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= In A- (Ea/RT) & R=8.31 x 10-³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/Rslope:- Ca = slope & - Ga=€agEa=-(-12571.42) x (8.3) x 10 Rg/mol-K)Ea=104.47 kj/md10.3- 14.1Y = (-14.1)-(-5.3) = -8.8x = (0.0037) - (0.0030) = 0.0007-8.80.0009Slope = rise=-12571.42sumIf we could extrapolate your graph to the Y-intercept, we would, knowing the shape of our graph and that theformula of straight line is y = mx + c; deduce the value of the slopey=mx+cInk = - € ax ( ² ) + &n A171816106From the shape of your graph, do you deduce/conclude that (circle your choice)a) the rate constant (decreases increases, or remains constant) with increasing activation energyb) the rate constant (decreases, (increases, or remains constant) with increasing temperature

According to Arrhenius equation, k = Ae-Ea/RT ln k =ln A -EaRT A plot of ln k versus 1/T gives a…

From the shape of your graph, do you deduce/conclude that (circle your chole a) the rate constant (decreases increases, or remains constant) with increasing activation energy b) the rate constant (decreases, (increases, or remains constant) with increasing temperature

From the shape of your graph, do you deduce/conclude that (circle your chole a) the rate constant (decreases increases, or remains constant) with increasing activation energy b) the rate constant (decreases, (increases, or remains constant) with increasing temperature

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

100%

1/T (K-¹)
2 N₂O5(g) → 4NO2(g) + O₂(g)
K(S-¹)
4.87 x 10-3
1.50 x 10-³
4.98 x 10-4
1.35 x 10-4
3.46 x 10-5
7.87 x 10-7
T (K)
338
0.0030
328
0.0030
318
0.0031
308
0.0032
298
0.0034
273
6.0037
The data in the table above provide the temperature dependence of the rate constant for the reaction
shown. Complete the table with required information to 2 significant figures each.
- a=-12571.42
In K, no units
5.3
- 6.5
7.6
8.9
Carefully construct a neat and well labelled graph/plot of In K (y-axis) versus 1/T (x-axis) in the appropriate
quadrant (s), using EXCEL or the template below; then determine the slope and activation energy, Ea for the given
reaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= In A- (Ea/RT) & R=
8.31 x 10-³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/R
slope:
- Ca = slope & - Ga=
€ag
Ea=-(-12571.42) x (8.3) x 10 Rg/mol-K)
Ea=104.47 kj/md
10.3
- 14.1
Y = (-14.1)-(-5.3) = -8.8
x = (0.0037) - (0.0030) = 0.0007
-8.8
0.0009
Slope = rise
=-12571.42
sum
If we could extrapolate your graph to the Y-intercept, we would, knowing the shape of our graph and that the
formula of straight line is y = mx + c; deduce the value of the slope
y=mx+c
Ink = - € ax ( ² ) + &n A
1718
16106
From the shape of your graph, do you deduce/conclude that (circle your choice)
a) the rate constant (decreases increases, or remains constant) with increasing activation energy
b) the rate constant (decreases, (increases, or remains constant) with increasing temperature

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