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"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

"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: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter4: Energy And Chemical Reactions
Section: Chapter Questions
Problem 4.DCP
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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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