Calculating an equilibrium constant from a partial equilibrium compositionAmmonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 1.5 L flask with3.5 atm of ammonia gas, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 2.1 atm.Calculate the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2significant digits.K = 0рx10☑

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Answered: Calculating an equilibrium constant…

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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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
Suppose a 250. mL flask is filled with 1.5 mol of CO, 0.80 mol of NO and 0.70 mol of CO,. The following reaction becomes possible: 2' NO, (g) +CO(g) - NO(g)+CO,(g) The equilibrium constant K for this reaction is 6.92 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. olo Ar OM ?
Suppose a 250. mL flask is filled with 0.70 mol of NO₂, 0.20 mol of NO and 1.9 mol of CO₂. The following reaction becomes possible: NO₂ (g) + CO (g) NO(g) + CO₂(g) The equilibrium constant K for this reaction is 0.715 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places.
Suppose a 500.mL flask is filled with 1.5mol of CO, 1.1mol of H2O and 1.6mol of H2. The following reaction becomes possible: CO(g)+H2O(g)+CO2(g)+H2(g) The equilibrium constant K for this reaction is 0.840 at the temperature of the flask. Calculate the equilibrium molarity of CO2. Round your answer to two decimal places.
Suppose a 250. mL flask is filled with 1.1 mol of Cl2, 0.20 mol of CHCl3 and 2.0 mol of HCl. The following reaction becomes possible: Cl2(g) + CHCl3(g) HCl(g) +CCI4(g) The equilibrium constant K for this reaction is 0.652 at the temperature of the flask. Calculate the equilibrium molarity of HCI. Round your answer to two decimal places. Ом X G
Suppose a 250. mL flask is filled with 1.6 mol of Br₂, 0.70 mol of OC12 and 0.50 mol of BrCl. The following reaction becomes possible: Br₂(g) + OC1₂(g) → BrOC1 (g) + BrCl(g) The equilibrium constant K for this reaction is 2.87 at the temperature of the flask. Calculate the equilibrium molarity of Br₂. Round your answer to two decimal places. M Ś
Suppose a 250. mL flask is filled with 1.7 mol of NO2, 0.80 mol of CO and 0.70 mol of CO2. The following reaction becomes possible: NO2(g) + CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 2.90 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.
True/False. Explain your answer in one sentence or less. In an equilibrium process, the concentrations of products and of reactants are equal. The value of the equilibrium constant for a given reaction depends on the initial concentrations of reactants. Large value for equilibrium constant K means the reaction will be less complete at the equilibrium point. The concentration of a pure solid is left out of an equilibrium constant expression but a pure liquid is included.
Suppose a 500. mL flask is filled with 0.50 mol of Cl₂, 1.7 mol of HCl and 0.20 mol of CC14. The following reaction becomes possible: Cl,(g)+CHCI, (g) → HCI(g) +CCI, (g) 4 The equilibrium constant K for this reaction is 0.104 at the temperature of the flask. Calculate the equilibrium molarity of Cl₂. Round your answer to two decimal places.
Consider the equilibrium system described by the chemical reaction below. Determine the concentration of O, at equilibrium by writing the equilibrium constant expression and solving it. Complete Parts 1-2 before submitting your answer. = 2 H₂O(g) 2 H2(g) + O̟₂(g) 1 2 NEXT At this temperature, the Kc = 2.4 × 103 and the equilibrium concentrations of H2O and H2 are 0.11 M and 0.019 M, respectively. If [x] represents the equilibrium concentration of O2, set up the equilibrium expression for Kc to solve for the concentration. Each reaction participant must be represented by one tile. Do not combine terms. Кс = 2' = 2.4 × 103 > RESET [0.11] [0.019] 2[0.11] 2[0.019] [0.11]² [0.019]² [x] [x]² [2x] [2x]²
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 22. mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 6.6 mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits. K-0 C = x10 X
Suppose a 250. mL flask is filled with 1.4 mol of CO, 1.9 mol of H₂O and 1.0 mol of CO₂. The following reaction becomes possible: CO(g) + H₂O(g) + CO₂(g) + H₂(g) 2 The equilibrium constant K for this reaction is 9.28 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. M X Ś ?

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