For the reaction,2N₂O5(g)The following data were obtained:T = 339 K[N₂O), M1.00 × 10Time0100300600900a. 45.8 kJ/molb.120.6 kJ/mol112.5 kJ/mol95.7 kJ/mol68.5 kJ/mol89.7 kJ/mol4NO2(g) + O2(g)C.d.e.f.*10²2.33 - 10-5.41- 10¹1.26 × 10¹XT = 310 K[N₂Os]. M1.00 × 109.54 × 10X8.63 10%7.43 - 106.39 × 10AT = 304 K[N₂O₂]. MFind the Ea using the best-fit straight line of the three rate constants at the three temperaturevalues.1.00 × 109.69 × 10²9.15 × 10²8.40 - 107.70 × 10²

Answered: Image /qna-images/answer/bca6ea63-6fa8-4a00-9fa0-a455808fa88f.jpg

For the reaction, 2N₂O5(g) The following data were obtained: T = 339 K [N₂O), M Time 0 100 300 600 900 4NO2(g) + O2(g) a. 45.8 kJ/mol b. 120.6 kJ/mol c. 112.5 kJ/mol d. 95.7 kJ/mol e. 68.5 kJ/mol 89.7 kJ/mol 1.00 - 10 6.14 10 2.33 - 10 5.41- 10¹ 1.26 × 10¹ X T = 310 K [N₂OJ. M 1.00 × 10 9.54 × 10² 8.63 10² 7.43 10 6.39 × 10 . T = 304 K [N₂O₂]. M 1.00×10 9.69 - 10 9.15 - 10 8.40 - 10 7.70 × 10² Find the Ea using the best-fit straight line of the three rate constants at the three temperature values.

For the reaction, 2N₂O5(g) The following data were obtained: T = 339 K [N₂O), M Time 0 100 300 600 900 4NO2(g) + O2(g) a. 45.8 kJ/mol b. 120.6 kJ/mol c. 112.5 kJ/mol d. 95.7 kJ/mol e. 68.5 kJ/mol 89.7 kJ/mol 1.00 - 10 6.14 10 2.33 - 10 5.41- 10¹ 1.26 × 10¹ X T = 310 K [N₂OJ. M 1.00 × 10 9.54 × 10² 8.63 10² 7.43 10 6.39 × 10 . T = 304 K [N₂O₂]. M 1.00×10 9.69 - 10 9.15 - 10 8.40 - 10 7.70 × 10² Find the Ea using the best-fit straight line of the three rate constants at the three temperature values.

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...

See similar textbooks

Similar questions

At 292.8 oC, the Kp for the reaction is 4.23⋅10-5: 3A+5B ↔ 1ABWhat is Keq?
Can you explain (not graded )
Estimate the temperature in K at which the reaction will become nonspontaneous.
← → C Secure https://www.webassign.net/web/Student/Assignment-Responses/submit?dep=29518359&tags-autosav Note: To find the value of the equilibrium constant at each temperature you must first find the value of AGO at each temp AGO=AHO - TASO For this reaction the values are ▲Hº = +88.3 kJ/mol and Asº= 151.3 J/mol*K 13. [0.05/0.1 Points] DETAILS 14. [0.1/0.1 Points] DETAILS Consider the reaction 2 NO2(g) → N₂O4(g) - (a) Using AGf N2O4(g) = 97.79 kJ/mol and AGF NO₂(g) = 51.3 kJ/mol, calculate AG at 298 K. -4.81 ✓ kj Consider the reaction given below. PREVIOUS ANSWERS (b) Calculate AG at 298 K if the partial pressures of NO₂ and N₂04 are 0.25 atm and 1.35 atm, respectively. 12.37 X KJ Course: 2022 Fall ENGL 200 X ILYEN DAD 2/4 Submissions Used PREVIOUS ANSWERS (a) Using thermodynamic data from the course website, calculate AG° at 298 K. -546.4 KJ 2/4 Submissions Used H₂(g) + F2(g) → 2 HF(g) E HE
*** S 19. Solid ammonium hydrogen sulfide is introduced into a 2.00-L flask, and the flask is şealed. If this solid decomposes according to the equation below DAUS Ca Nothange NH4HS(s) NH3(g) + H₂S(g), Kp = 0.108 at 25°C, what is the minimum mass of ammonium hydrogen sulfide that must be present in the flask initially if equilibrium is to be established at 25°C? A) 0.917 g B) 1.37 g C) 2.74 g D) 0.581 g E) 0.452 g
For the reaction: 2SO2(g) + O2 (g) ⇄ 2SO3(g), ∆H = 102.4kJ. Indicate how each of the listed stressors willaffect the item listed in the column header.
Consider the following equilibrium: N204(g) = 2NO2(g) Suppose two different experiments were conducted on this equilibrium, at two different temperatures. Trial Temperature Initial Moles of N204 Equilibrium Moles of N204 1 25°C 1.00 0.815 55°C 1.00 0.500 What conclusion may be drawn from the above data? O Kc> Kp at all temperatures. The reaction is exothermic. The forward reaction proceeds faster than the reverse reaction. OKp =D1 at 55°C. O The equilibrium constant increases with increasing temperature.
Calculate the value of K, for the reaction 2N₂(g) + O₂(g) 2N₂O(g) at 298.15 K and 1473 K. Thermodynamic data for N₂O(g) are: AH = 82.05 kJ/mol; Sº = 219.9 J/mol -K; AG% = 104.2 kJ/mol. 1st attempt Part 1 2 298.15 K Part 2 1473 K
Determine how the following changes affect the given equilblirium reaction that is held in a closed container. A(g) + 2 B(s) ⇌ 3 C(g) + 3 D(s) ΔH°rxn = −0.25 kJ/mol Adding some A(g) Adding some D(s) Removing some C(g) Increasing the temperature Halving the volume of the container The following mechanism is proposed for the reaction 2 A2 + B2 → 2 C A2 + B2 → R + C slow A2 + R → C fast What is the rate law for the overall reaction? A. rate = k[A2] B. rate = k[A2]2 [B2] C. rate = k[A2]2 D. rate = k[R]2 E. rate = k[A2][B2]