Answered: Calculate the activation energy for the…

Calculate the activation energy for the reverse process in kJ. Ep Products AE = 520 kJ AE = 180 kJ %3D Reactants Reaction Path hemical Thinking nnn,

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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|For the equilibrium system CuO (s) + CO (g) = Cu (s) + CO2 (g), predict the effect that the indicated stress will have on the specified quantity. a) The effect of adding CO on the amount of Cu (s) b) The effect of adding CO2 on [CO] c) The effect of removing CuO on the [CO2] d) The effect of a catalyst on the amount of Cu (s) Do any of the stresses affect the value of K?
Using the potential energy diagram determine the activation energy (each division is 10kg).
reaction CH;OH + Br– C‚H;Br + OH is shown below: E (kJ) C2HSBR+ OH 161 71 CaHOH + Br Reaction Coordinate Which of the following are true? Choose all that apply. The magnitude of Ea for the reverse reaction is smaller than 161 kJ. O The energy of the products is higher than the energy of the reactants. O If the energy of the activated complex were increased, AE would increase. O AE is negative.
[Review Topics] [References] Use the References to access important values if needed for this question. A reaction profile (not to scale!) for the reaction C₂H4+ HC1 C₂H5 Cl is shown below: E (KJ) C₂H4 + HCI 70 205 Reaction Coordinate C₂HCI Which of the following are true? (Select all that apply.) OThe magnitude of Ea for the reverse reaction is larger than 205 kJ. Othe energy of the products is lower than the energy of the reactants. OThe value of AE in the presence of a catalyst would be smaller than -70.0 kJ. OAE is positive. OThe value of Ea in the presence of a catalyst would be larger than 205 kJ.
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Question Completion Status: QUESTION 7 The following reaction is exothermic: CF2C12(g) CF2CI(g) + Cl(g) Select the energy diagram that correctly shows the relative energies of the reactants, products, and the activated complex as well as the correct molecular representations of reactants, products, and possible structure for the activated complex. Eproducts E. EReactants Reaction Progress EProducts E, EReactants Reaction Progress Eproducts Ea EReactants Reaction Progress EProducts Ea EReactants Save All Arnswers Reaction Progress Click Save and Submit to save and submit. Click Save All Answers to save all answers. Energy Energy Energy
← → 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
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Hi i am having trouble with this portion of the question. Please provide handwritten solution for better understanding CO(g) + Cl2(g) COCl2(g) The above reaction obeys the mechanism: Cl2 --> 2Cl Fast equilibrium Cl + CO--> COCl Fast equilibrium COCl + Cl2 --> COCl2 + Cl Slow 2Cl--> Cl2 Fast 1b) Suppose that: k1/k-1 = 1×10-2 mol L-1, k2/k-2 = 1×106 L mol-1, k3 = 1×10-4 L mol-1 s-1, and k4 = 1×103 L mol-1 s-1. What would be the value of the overall rate constant, k, including the correct units?
Consider the quilibrium reaction between X and Y, as shown below: X=Y AG The reaction is started with 10 mmol of X; no Y is initially present. After 48 hours, analysis reveals the presence of 10 mmol of X and 0 mmol of Y. Which is the most likely explanation? = −1 - 45 kJ mol X and Y have reached equilibrium concentrations. An enzyme has shifted the equilibrium toward X. Formation of Y is kinetically slow; equilibrium has not been reached by 48 hours. Formation of Y is thermodynamically unfavorable. Two of the above explanations are reasonable.
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.