12. In the Ostwald process for oxidizing ammonia, a variety of products is possible - N2, N₂O, NO, and NO₂-depending on the conditions. One possibility is:NH3(g) + O₂(g) = NO(g) + H₂O (g) Kp = 2.11 x 10¹⁹ at 700 KFor the decomposition of NO2 at 700 K,NO₂(g) = NO(g) + O2(g) Kp = 0.524 at 700 KWhat is Kp for the oxidation of NH3(g) to NO2(g)?13. The decomposition of N₂O4 to NO₂ is represented by the equilibrium reaction equation belowN2O4(g) 2 NO2 (g)If 1.75 moles of N₂O4 is injected in a 1.5 L vessel and at equilibrium 25% N₂O4 is dissociated, what is theequilibrium constant for this reaction?

Answered: Image /qna-images/answer/5138ca45-729f-4605-ba08-342c3f952db7.jpg

12. In the Ostwald process for oxidizing ammonia, a variety of products is possible - N₂, N₂O, NO, and NO₂- depending on the conditions. One possibility is: NH3(g) + O₂(g) = NO(g) + H₂O (g) Kp = 2.11 x 10¹⁹ at 700 K For the decomposition of NO2 at 700 K, NO2(g) = NO(g) + O2(g) Kp = 0.524 at 700 K What is K, for the oxidation of NH3(g) to NO2(g)? 13. The decomposition of N₂O4 to NO₂ is represented by the equilibrium reaction equation below N2O4(g) 2 NO2 (g) If 1.75 moles of N₂O4 is injected in a 1.5 L vessel and at equilibrium 25% N₂O4 is dissociated, what is the equilibrium constant for this reaction?

12. In the Ostwald process for oxidizing ammonia, a variety of products is possible - N₂, N₂O, NO, and NO₂- depending on the conditions. One possibility is: NH3(g) + O₂(g) = NO(g) + H₂O (g) Kp = 2.11 x 10¹⁹ at 700 K For the decomposition of NO2 at 700 K, NO2(g) = NO(g) + O2(g) Kp = 0.524 at 700 K What is K, for the oxidation of NH3(g) to NO2(g)? 13. The decomposition of N₂O4 to NO₂ is represented by the equilibrium reaction equation below N2O4(g) 2 NO2 (g) If 1.75 moles of N₂O4 is injected in a 1.5 L vessel and at equilibrium 25% N₂O4 is dissociated, what is the equilibrium constant for this reaction?

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

11
At 100°C, Kₚ=60.6 for the reaction 2NOBr(g)⇌ 2NO(g)+Br₂(g) In a given experiment, 0.10 atm of each component is placed ina container. Is the system at equilibrium? If not, in which direc-tion will the reaction proceed?
The value of Kp for the reaction 2 A(g) + B(g) + 3 C(g) → 2 D(g) + E(g) is 16870 at a particular temperature. What does the magnitude of Kp tell us about the equilibrium position of the reaction? A) There are many more products than reactants. B) There are only a few more products than reactants. C) There are many more reactants than products. D) There are only a few more reactants than products.
Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet proof glass. One of the compounds used to make Lexan is phosgene, COCI2, an extremely poisonous gas. Phosgene decomposes by the reaction: CoCl₂ (g) CO (g) Cl2 (g) For this reaction Kp = 6.8x109 at 100°C. If phosgene is at an initial pressure of 1.097 atm, calculate the equilibrium pressures of all species.
11. The reaction below is prepared with an initial pressure of hydrogen sulfide, H2S, of 2.00 atm. At equilibrium you determine that the partial pressure of S2 is 0.25 atm. Determine Kp for the reaction. 2H2S(g) = 2H2(g) + S2(g)
Consider the reaction: 2HI(g) ⇀ H2(g)+I2(g). At 500 C∘, a flask initially has all three gases, each at a partial pressure of 0.200 atm. When equilibrium is established, the partial pressure of HI(g) is determined to be 0.480 atm. a. Calculate the Kp for the reaction b. What is the equilibrium partial pressure of H2(g) and I2(g)? c. What is ΔG⊖reaction?
The value of Kp for the reaction 2 A(g) + B(g) + 3 C(g) → 2 D(g) + E(g) is 12770 at a particular temperature. What does the magnitude of Kp tell us about the equilibrium position of the reaction? A) There are many more products than reactants. B) There are only a few more products than reactants. C) There are many more reactants than products. D) There are only a few more reactants than products.
A student ran the following reaction in the laboratory at 583 K: COCI₂(9) co(g) + Cl₂(9) When he introduced COCI₂(g) at a pressure of 0.417 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of Cl₂(g) to be 0.243 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp = =
Some quantity of NOBr is added to an otherwise empty flask. The reaction: 2 NOBr (g) ⇄ 2 NO (g) + Br2 (g) takes place at a particular temperature for which KP is 3.5 x 10-5. Assuming an equilibrium pressure of NOBr = 2.46atm, what is the equilibrium pressure of NO?
4. IBr decomposes according to the reaction at 150 C: 2 IB (g) = 12 (g) + Brz (g) If IBr is placed in a container at an initial pressure of 5.0 atm, what will be its equilibrium pressure at 150°C? K = 8.3 x 10-3
Carbon tetrachloride can be produced by the following reaction: CS₂(g) + 3 Cl₂ (g) S₂Cl₂(g) + CCL (g) Suppose 1.20 mol of CS₂ and 3.60 mol of Cl₂ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.23 mol CCl4. Calculate Ke at the unknown temperature. Ke- Submit Answer Try Another Version References] 10 item attempts remaining
The reaction mixture below contains 0.305 atm of H2, 0.060 atm of S2, and 0.420 atm of H2S. 2 H2S(g) 2 2 H2(g) + S2(g) Kp = 2.4x104 at a particular temperature. Based on the information provided, select the correct relationship. K; the reaction will proceed to form more S2(g). Q > K; the reaction will proceed to the left. O More experimental data are needed to answer a question of this type. Q = K; the reaction is at equilibrium and no change is expected.