C doublingthevo lumeofthecontainer holdingthemixture

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Unformatted text preview: 1)(400)]1 = 159 ˆ ˆˆ HCONH2 (g) ‡ˆ† NH3 (g) + CO (g) Initial 2.83 atm 0 0 – x + x + x D Equilibrium 2.83 – x x x K p = PNH 3 P O C PHCONH 2 = 2 (x) = 159 (2.83 ­ x) Solving th e quadratic equation, x = 2.78 PHCONH 2 = 2.83 ­ 2.78 = 0.05 atm PNH 3 = PCO = 2.78 atm PTotal = 2.78 + 2.78 + 0.05 = 5.61 atm 4 10. Predict how each of the fo llowing changes affect the amount of H2 present in an equilibrium mixture in the reaction ˆˆ 3 Fe (s) + 4 H2O (g) ‡ˆ† Fe3O4 (s) + 4 H2 (g) ∆H = −150 kJ a) Raising the temperature of the mixture. Since reaction is exothermic, raising temperature will shift the equilibrium to the left (¬) and reduce amount of hydrogen. b) Adding more H2O (g). Adding more water, will shift the equilibrium to the right (®) and increase the amount of hydrogen. c) Doubling the vo lume of the container holding the mixture. Increasing the volume of the container will reduce the pressure but the equilibrium will not be affected, and the amount of hydrogen will not change. d) Adding a catalyst. Adding catalyst does not alter the equilibrium and the amount of hydrogen produced. 3 11. At 2000 °C the equilibrium constant for the reaction below is Kc= 2.4x10 . If the init ial concentration of NO is...
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This document was uploaded on 03/18/2014 for the course CHEM 102 at Los Angeles Mission College.

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