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CH302 Chapter 6 answers part 3

# CH302 Chapter 6 answers part 3 - Use of Kc to find...

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Use of K c to find Equilibrium Concentrations (1) K c , = 3.00 for the following reaction at a given temperature. If 1.00 mole of SO 2 and 1.00 mole of NO 2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium? +, M x M x M x M x M x M x M x M x M M 500 . 0 500 . 0 m Equilibriu + + - - Change 0 0 0.500 0.500 Initial NO SO NO SO (g) 3(g) 2(g) 2(g) 0 0 m. o . The concentrations of SO 3 and NO are zero, making Q zero so obviously the forward reaction must occur. . At equilibrium, x moles/L of SO 2 and NO 2 will have reacted… forming x moles of SO 3 and NO Use of K c to find Equilibrium Concentrations (1) >@ > @ > @ equation. the sidesof both of the can take We . square perfect a is equation This 500 . 0 500 . 0 00 . 3 NO SO NO SO K 500 . 0 500 . 0 m Equilibriu + + - - Change 0 0 0.500 0.500 Initial NO SO NO SO 2 2 3 c (g) 3(g) 2(g) 2(g) x x x x M x M x M x M x M x M x M x M x M M 0 0 0 0 o . We know that K c = 3.0 so we write out the full expression and substitute in the concentrations: This leaves us with 1.73 = x/ (0.500- x) which solves to give x = 0.316

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Use of K c to find Equilibrium Concentrations (1) Final answer: using x = 0.316 +, M x M x M x M x M x M x M x M x M M 500 . 0 500 . 0 m Equilibriu + + - - Change 0 0 0.500 0.500 Initial NO SO NO SO (g) 3(g) 2(g) 2(g) 0 0 m. o . 0. 184M 0.184M 0.316M 0.316M Use of K c to find Equilibrium Concentrations (2) K = 49 for the following reaction at 450 o C. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance? m o (g) 2(g) 2(g) HI 2 I + H The concentrations of H 2 and I 2 are zero, making Q infinite which is bigger than 49!! We can see this is obviously not at equilibrium and some of the reverse reaction must occur. At equilibrium, x moles/L of HI will have reacted… forming x/2 moles of H 2 and I 2 More simply, 2x moles/L of HI will have reacted… forming x moles of H 2 and I 2 Use of K c to find Equilibrium Concentrations (2) >@ > @ > @ M M x M M x M x x x x x x x x x M x M x M x M x M x M x M 78 . 0 2 00 . 1 HI 11 . 0 I H 11 . 0 ; 00 . 1 9 ; 2 00 . 1 0 . 7 2 - 1.00 = 7.0 = K 2 - 1.00 = 49 = I H HI = K 2 - 1.00 m Equilibriu 2 - + + Change 1.00 0 0 Initial HI 2 I + H 2 2 c 2 2 2 2 c (g) 2(g) 2(g) 0 0 m o Calculating an Equilibrium Constant (1) One mole of PCl 5 was introduced into an evacuated 1.00 liter container at some temperature. The system was allowed to reach equilibrium. At equilibrium 0.60 mole of PCl 3 was present in the container. Calculate K at this temperature. PCl 5(g) PCl 3(g) + Cl 2(g) Initial 1.00 M 0 M 0 M Change -0.60 M +0.60 M +0.60 M Equilibrium 0.40 M 0.6 M 0.6 M So K c = [PCl 3 ] [Cl 2 ] / [PCl 5 ] = ( 0.6 * 0.6 ) / 0.4 = 0.90
Calculating an Equilibrium Constant (2) 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 .

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CH302 Chapter 6 answers part 3 - Use of Kc to find...

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