Chapter 17!!

Chapter 17!! - Chapter 17: Chemical Equilibrium A+BC+D A+ B...

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Chapter 17: Chemical Equilibrium A + B → C + D A+ B ↔ C + D -Reaction is reversible -forward and reverse reaction have a rate -dynamic equilibrium -total concentration become fixed -but reactants still form products, etc. -stressor -impact the forward or reverse reaction -then continue to react to establish new equilibrium [A] = number – x [B] = x K = x/ number – x Equilibrium Constant A (g) + B (g) →← C (g) + D (g) k f /k r =K c = [C][D]/[A][B] Example 17-2 : One liter of equilibrium mixture from the following system at a hight temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate K c for the reaction. PCl 5 ↔ PCl 3 + Cl 2 I C E 0.028 M 0.172 M 0.086 M K c = [PCl 3 ][Cl 2 ]/[PCl 5 ] K c = (0.172M)(0.086M)/(0.028M) K c = 0.528 Example 17-3 : The decomposition of PCl 5 was studied at another temperature. One mole of PCl 5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl 3 was present in the container. Calculate the equilibrium constant at this temperature. PCl 5 ↔ PCl 4 + Cl 2 I 1 mol 0 mol 0 mol C -0.60 mol +0.60 mol +0.60 mol E 0.40 mol 0.60 mol 0.60 mol K c = 0.9 Example 17-4 : At a given temperature 0.80 mole of N 2 and 0.90 mole of H 2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH 3 was present. Calculate K c for the reaction. N 2 + 3H 2 ↔ 2NH 3 I 0.8M 0.9 M 0 M C -0.1M -0.3 M +0.2M E 0.7 M 0.6 M 0.2 M K c = 0.265 Example 17-5 Q = K equilibrium
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Example 17-7 : The equilibrium constant, K c , is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO 2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at
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Chapter 17!! - Chapter 17: Chemical Equilibrium A+BC+D A+ B...

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