13. The equilibrium constant, Ke, for the following reaction is 54.3 at 430 °C. H2(g) + 12 (g) = 2HI (g) At the start of the reaction there are 0.714 mole of hydrogen, 0.985 mole of iodine, and 0.886 mole of hydrogen iodide in a 2.40 L reaction chamber. Set up an ICE table and describe the steps needed to determine the concentrations of the gases at equilibrium. You do not need to solve the equation.

13. The equilibrium constant, Ke, for the following reaction is 54.3 at 430 °C. H2(g) + 12 (g) = 2HI (g) At the start of the reaction there are 0.714 mole of hydrogen, 0.985 mole of iodine, and 0.886 mole of hydrogen iodide in a 2.40 L reaction chamber. Set up an ICE table and describe the steps needed to determine the concentrations of the gases at equilibrium. You do not need to solve the equation.

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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Daniela A chemical engineer is studying the following reaction: 2 H,S(g)+302(g) → 2 SO2(g)+2H,O(g) At the temperature the engineer picks, the equilibrium constant K for this reaction is 2.0. The engineer charges ("fills") four reaction vessels with hydrogen sulfide and oxygen, and lets the reaction begin. She then measures the composition of the mixture inside each vessel from time to time. Her first set of measurements are shown in the table below. Predict the changes in the compositions the engineer should expect next time she measures the compositions. reaction vessel compound expected change in pressure pressure H, S 1.52 atm O t increase O I decrease O (no change) 0, 8.80 atm O t increase O I decrease O (no change) A so, 9.53 atm O t increase O I decrease O (no change) H,0 5.94 atm O f increase O I decrease O (no change) H, S 0.79 atm O t increase O I decrease O (no change) 6.91 atm Ot increase O l'decrease O (no change) so, 8.61 atm O t increase O I decrease O (no change) H,0…
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