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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
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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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
Transcribed Image Text: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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