"if we could extrapolate your graph to your Y-intercept, we would, knowing the shape of our graph and that the formula of straight line is y=max+c, deduce the value of slope

Chemistry
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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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I don't understand what I have to demonstrate on the question about "if we could extrapolate your graph to your Y-intercept, we would, knowing the shape of our graph and that the formula of straight line is y=max+c, deduce the value of slope 

T (K)
338
328
318
308
298
273
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
In K, no units
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.
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, E, for the given
reaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= ln A- (E₂/RT) & R=
8.31 x 10³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/R
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
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
Transcribed Image Text:T (K) 338 328 318 308 298 273 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 In K, no units 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. 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, E, for the given reaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= ln A- (E₂/RT) & R= 8.31 x 10³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/R 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 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
In K
5
0
-2
-4
-6
-10
-12
-16
In K vs 1/T
0.0005 0.001 0.0015
1/1
0.0025
0.003
0.0035 0.004
y=-10696x+25.624
M
Transcribed Image Text:In K 5 0 -2 -4 -6 -10 -12 -16 In K vs 1/T 0.0005 0.001 0.0015 1/1 0.0025 0.003 0.0035 0.004 y=-10696x+25.624 M
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