QUESTION 26 For the gas phase reaction below, AH = -1.6 102 kJ for the forward reaction. SO2 + ½ 02SO3 In order to increase the yield of SO3, the reaction should be run O a. at low P, low T. b. at high P, high T. O c. at low P, high T. O d. at high P, but is independent of T. O e. at high P, low T.

QUESTION 26 For the gas phase reaction below, AH = -1.6 102 kJ for the forward reaction. SO2 + ½ 02SO3 In order to increase the yield of SO3, the reaction should be run O a. at low P, low T. b. at high P, high T. O c. at low P, high T. O d. at high P, but is independent of T. O e. at high P, low T.

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

Part A The reaction below is carried out at a different temperature at which Ke = 0.055. This time, however, the reaction mixture starts with only the product, NO = 0.0100 M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium. N2(8) + O2(g) = 2NO(g) Express your answers to three significant figures separated by commas. > View Available Hint(s) ? [N2]. [O2]. [NO] = М, М, М Submit Provide Feedback Next >
8. NO, is soluble in carbon tetrachloride (CC4). As it dis- solves, it dimerizes to give N,O4 according to the equation 2 NO2(CC1) 2 N,O4(CC14) A study of this equilibrium gave the following experimen- tal data at 20°C: [N,O,] (mol L-1) [NO,] (mol L-1) 0.192 x 10-3 2.68 x 10-3 0.721 x 10-3 4.96 x 10-3 1.61 x 10-3 7.39 x 10-3 2.67 x 10-3 10.2 x 10-3 3.95 x 10-3 11.0 x 10-3 7.90 x 10-3 16.6 x 10-3 11.9 x 10-3 21.4 x 10-3 (a) Graph the concentration of N,O4 versus the square of the concentration of NO2. (b) Compute the average equilibrium constant of this reaction.
20 <. At a certain temperature, 0.940 mol SO, is placed in a 2.50 L container. 2 SO, (g) =2 SO, (g) + 0,(g) At equilibrium, 0.140 mol O, is present. Calculate Ke. K. = 2.7738 x10-2 Incorrect help about us privacy policy terms of use careers contact us O Bi w
10 For a chemical reaction: Cl2 (1) 0.55 mol Cl₂ Solve for the final partial pressure of Cl in atm: A. 1.8 × 108 B. 8.9 x 105 D. 3.6 × 104 2 Cl (g) has a Kp =3.2 x 10", and the initial amounts are: 0 atm Cl E. 0.55 C. 1.8 104 10.8
Please help, thank you!
What is the equilibrium concentration for the product (P) in the reaction R ---> P if the DeltaGo' = +5.7 kJ/mol and initial concentrations of P and R are both 1.00 M? a. 0.18 M b. 0.10 M c. 1.82 M d. 10.0 M e. none of the above
Picture
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.
Given this reaction at equilibrium, C(s) + CO2(g) D 2CO(g) (∆H° = 119kJ), explain the changes that would occur when the following stresses are applied: a.CO is removed. b.Heat is removed. c.CO2 is added.
A. Consider the endothermic reaction at 25oC: FeSCN+2(aq) --><-- Fe+3(aq) + SCN-(aq) What direction will the reaction shift in response to the following stresses? (Use arrows or write "right" , "left" , or "no shift"). a. More SCN- is added. b. Solid sodium hydroxide is added. (iron (III) hydroxide is insoluble). c. The reaction container is set into ice water to cool. d. The pressure of the air around the reaction is doubled. e. Solid iron (III) chloride (FeCl3) is added. B. At 25oC, Kc for thr above reaction is 9.1 x 10-6. Determine the equilibrium concentration of Fe+3 in the solution if 0.50 mol of FeSCN+2 is dissolved to make 2.50 liters of solution.