Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene, COCl₂, an extremely poisonous gas. Phosgene decomposes by the reaction:\[ \text{COCl}_2 \, (\text{g}) \rightleftharpoons \text{CO} \, (\text{g}) + \text{Cl}_2 \, (\text{g}) \]For this reaction, \( K_p = 6.8 \times 10^{-9} \) at 100°C. If phosgene is at an initial pressure of 1.097 atm, calculate the equilibrium pressures of all species.

Consider the given equilibrium equation is as follows; COCl2 g ↔ COg + Cl2g…

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: COCI2 (g) + CO (g) Cl2 (8) 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.

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: COCI2 (g) + CO (g) Cl2 (8) 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.

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 a certain temperature, the Kp for the decomposition of H2S is 0.886. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.228 atm in a closed container. What is the total pressure in the container at equilibrium?
2HOCI(g) ⇒ H₂O(g) + Cl₂O(g) Kp = 11.1 @ 298 K The reaction begins with 2.5 atm of HOCI, 0.5 atm H₂O, and 1.0 atm Cl2O. Calculate the equilibrium pressures of the components. What is the equilibrium pressure of HOCI? Рносі = ?] atm
Phosphine gas (PH3), decomposes by the reaction shown below. A total pressure of 0.64 atm of pure phosphine gas was placed in a sealed 1.00-L flask at 298 K. After equilibrium was attained, the total pressure inside the flask was found to be 0.93 atm. Calculate the mass of solid phosphorous in the flask. PH3(g) <--> P(s) + 3 H2(g)
EXPLAIN STEP BY STEP
10) For the reaction: 2 V PC13(g) + Cl2(g) → PC15(g) at 85°C, Kp = 1.19 = If one starts with 2.00 atm pressure of PC13, 1.00 atm pressure of Cl2 and no PC15, what is the partial pressure of PC15(g) at equilibrium? A) 0.739 atm B) 0.621 atm C) 0.465 atm D) 0.167 atm E) 0.553 atm
Matak
The elementary reaction 2 H, O(g) = 2 H, (g) + 0, (g) proceeds at a certain temperature until the partial pressures of H,O, H,, and O, reach 0.0350 atm, 0.00650 atm, and 0.00150 atm, respectively. What is the value of the equilibrium constant at this temperature? Kp
Nitrosyl chloride decomposes significantly only at elevated temperatures. At the room temperature, its decomposition is characterized by a small value of equilibrium constant: 2NOCl(g) ⇄ 2NO(g) + Cl2(g), Kp = 7.0 \times 10-8 atm at 250C Suppose you fill a container with 1.00 atm of pure NOCl (g) at 25℃ and wait until the decomposition equilibrium is established. What equilibrium partial pressures of NOCl(g), NO(g), and Cl2(g) are expected? Do not give handwriting solution.
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.
At 1560 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP = 0.846. If the initial pressure of IBr is 0.00674 atm, what are the equilibrium partial pressures of IBr, I2, and Br2?p(IBr) = p(I2) = p(Br2) =
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
At 150°C, Kp = 1.2x10- for the reaction 12(g) + Br2(g) 5 2Brl(g). A rigid cylinder at that temperature contains 0.127 atm of bromine, 0.134 atm of iodine, and 1.055 atm of bromine monoiodide. What is the value of the reaction quotient, Qp? O 55.2 62.0 O 65.4 O 88.3