CO(g) + 2H₂(g) CH₂OH(g) The reaction between CO and H₂ is carried out at a specific temperature with initial concentrations of CO = 0.26 M and H₂ = 0.55 M. At equilibrium, the concentration of CH, OH is 0.12 M. Part A Find the equilibrium constant at this temperature. Express your answer using two significant figures. for Part for Pardo for Part redo foPart A resor Part A keyboard shortcuts for Part A help for Part A Ke 15 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

CO(g) + 2H₂(g) CH₂OH(g) The reaction between CO and H₂ is carried out at a specific temperature with initial concentrations of CO = 0.26 M and H₂ = 0.55 M. At equilibrium, the concentration of CH, OH is 0.12 M. Part A Find the equilibrium constant at this temperature. Express your answer using two significant figures. for Part for Pardo for Part redo foPart A resor Part A keyboard shortcuts for Part A help for Part A Ke 15 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

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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Knowing that [reactant]Ó = [reactant]initial - [reactant]onsumed And that Fe"(aq) + SCN'(aq) Ó Fe(SCN)+*(aq) Complete the table below and determine an average value for the equilibrium constant K. Exp#1 exp#2 exp#3 [Fe*1° (M) 1.65 x 10 1.65 x 103 1.65 x 103 [SCN]° (M) 3.30 x 10 6.60 x 10 9.90 x 104 [Fe(SCN)*jÓ (M) 9.00 x 105 1.78 x 10 2.67 x 10 [Fe*jó ISCNJÓ K Average K = Fe"(aq) + SCN (aq) ó Fe(SCN)**(aq) I 1.65 x 10 -9.00 x 10° 3.30 x 10 -9.00 x 105 C 0.00 +9.00 x 10 9.00 x 10$
The reaction quotient for a system is 3.9 x 103. If the equilibrium constant for the system is 2 x 10-4, what will happen as the reaction mixture approaches equilibrium? O a) There will be a net gain in reactant(s). O b) There will be a net gain in product(s). O C) The equilibrium constant will increase until it equals the reaction quotient. O a) The equilibrium constant will decrease until it equals the reaction quotient. e) There will be a net gain in both product(s) and reactant(s).
E8hstant from a heterogeneous. Iron(III) oxide and hydrogen react to form iron and water, Ilike this: Fe,0;(s)+3 H,(9)→2 Fe(s)+3H,O(g) At a certain temperature, a chemist finds that a 4.0 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has following composition: compound amount Fe,O3 4.85 g H2 2.01 g Fe 3.53 g H,O 1.33 g Calculate the value of the equilibrium constant K, for this reaction. Round your answer to 2 significant digits. K = | Explanation Check 2022 McGraw H LLC A Rights Reservedt Tem of Use acy Conter Accesbity Type here to search 99 C FIO Pr Sc 24 4. & he 7. 8. %3D Y U H K Alt Alt Chl
Consider the reaction. NaC2O2H3(s) <--- H2O ------> NaC2O2H3(aq), which is at equilibrium in an open flask in the lab at room temperature. You add more water to the equilibrium mixture. When the system reestablishes equilibrium, what has changed? a. the concentration of NaC2H3O2(aq) does not change b. heat is produced c. the concentration of NaC2H3O2(aq) decreases d. the concentration of the water increases
1. The reaction shown below is carried out at various temperatures and the value of Kc determined. Draw particulate representations that show the relative amounts of each reactant and product that would roughly correspond to the value of K. Legend = PC13 = Cl₂ ∞ = = PC15 PC13 (g) + Cl2 (g) = PC15 (g) Magnitude of K Particulate diagram of equilibrium mixture 1 x 10 -3 1 x 105
Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: 2 HNO3(aq)+NO(g)→ 3 NO,(g)+H,0(1) At a certain temperature, a chemist finds that a 4.5 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount dlo HNO3 14.2 g 18 Ar NO 7.3 g NO, 19.2 g H,O 87.0 g Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits. K = 0 x10 ?
5. The reaction: 2NO₂(g) + 7H₂(g) = 2NH3(g) + 4H₂O(g) was allowed to reach equilibrium at 455 K. Analysis showed that there was 34.6 g of NO2, 53.2 g of H₂, 52.6 g of NH3 and 87.4 g H₂O present in a 2.65 L vessel once equilibrium was achieved. a. Determine the equilibrium concentration for both reactants. b. Determine the equilibrium concentration for both products. C. Write the Kc expression. d. Solve for the Kc. e. Solve for the Kp at the same temperature.
What is the equilibrium concentration of BCl3 if a solid sample of PH3BC13 is placed in a 0.250 L closed vessel at 80.0 °C and decomposes until equilibrium is reached? Assume that there is solid present at equilibrium. At 80.0 °C, Kc = 1.87 x 10-³ for the reaction: PH3BC13(s) ⇒ PH3(g) + BCl3(g) Provide your answer in M, without units, and use the correct number of significant figures. 534.8
For the generic reaction, A (g) = B (g) consider each value of K and initial concentration of A. For which set of values does the x is small approximation most likely apply? ОК - 1.0 х 106, [A] - 0.150 М K= 1.0 x 10-1, [A] = 0.150 M O K= 1.0 x 106, [A] = 0.00150 M ок-1.0 х 101, [A] - 0.00150 М
Suppose a 500. mL flask is filled with 1.3 mol of N₂ and 1.4 mol of O₂. The following reaction becomes possible: N₂(g) + O₂(g) → 2NO(g) The equilibrium constant K for this reaction is 7.02 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. M X S
Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.2 x 104 at a certain temperature. If a solid sample of NH.SH decomposes, what will the equilibrium concentration of NH3 be? NH:SH(s) = NH3(g) + H2S(g) 1 2 3 NEXT > Based on the given values, set up ICE table in order to determine the unknown. NH.SH(s) NH:(g) H2S(g) Initial (M) Change (M) Equilibrium (M) RESET 1.2 x 104 +x +2x -2x 1.2 x 104 + x 1.2 x 104 - x 1.2 x 104 + 2x 1.2 x 104- 2x
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