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Chem102-Chap_14

# 387mo lh2o calculatetheequilibriumconstantfor

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Unformatted text preview: g) ¾ Kc = Example 2: The Equilibrium Constant for a reaction is: K c = [NH 3 ]4 [O 2 ]5 4 [NO] [H 2 O]6 What is the Equilibrium Constant expressio n when the equation for this reaction is halved and then reversed? Original Equat ion: ¾¾ ® 4 NO + 6 H2O ¬¾ 4 NH3 ¾ + 5 O2 Halved and Reversed: Kc = 6 Chemistry 102 Chapter 14 CALCULATING KC FOR REACTIONS · Molar concentrations (mo larit ies) of products and reactants at equilibrium must be subst ituted in the expressio n of Kc Example 1: CO (g) + ¾¾ ® 3 H2 (g) ¬¾ ¾ CH4 (g) + H2O (g) When 1.00 mo l CO and 3.00 mol H2 are placed in a 10.0 L vessel at 927 °C and allowed to come to equilibrium, the mixture is found to contain 0.387 mo l H2O. Calculate the equilibrium constant for this react ion. ¾¾ ® CO (g) + 3 H2 (g) ¬¾ CH4 (g) + H2O (g) ¾ Init ial 1.00 mo l 3.00 mo l 0 0 D – 0.387 – 1.16 + 0.387 + 0.387 Equilibrium 0.613 mol 1.84 mol 0.387 mol 0.387 mol CO: H2 : CH4: H2O: Equilibrium amounts: 0.613 mo l 1.84 mo l 0.387 mo l 0.387 mo l K c = Equilibrium Concentrations(M) 0.613 mo l/10.00...
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