Formation of Precipitate and Ke to Kp

Formation of Precipitate and Ke to Kp - The relationship...

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1 L of 0.02 M CaCl 2 and 1 L of 0.0004 M Na 2 SO 4 are mixed together. Will there be a precipitate of CaSO 4 formed in this reaction? (no precipitate is formed.) 2. 500 mL of 4 x 10 -5 M AgNO 3 is mixed with 500 mL 8 x 10 -4 M NaCl. Will precipitate form? (Yes) Solution According to the solubility table, NaNO 3 is soluble and AgCl can form a precipitate. So you will be proceeding with AgCl 3) 100 mL of 2 x 10 -2 M Na 2 CrO 4 is mixed with 100 mL 4 x 10 -3 M Ba(NO 3 ) 2 . Predict the formation of precipitate. (Yes)
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1. In an equilibrium system involving gases, the equilibrium constant can also be expressed with the partial pressures of the gases. The designation for the equilibrium constant, when the partial pressures are shown in atmospheres, is Kp. For the reaction, 2NO(g) + O2 2NO2(g) the equilibrium constant, Kp, for the reaction would be: If molar concentrations are given the equilibrium constant expression would be
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Unformatted text preview: The relationship between Ke and Kp is the following: Kp = Ke(0.0821x T ) Δn T represents the temperature in Kelvins. Δ n = Change in total number of gaseous moles from the balanced stoichiometric equation ( n products – n reactants). The 0.0821 is the gas constant, R . The units are L-atm/mol-K. Example For the general equilibrium reaction, A(g) + 2B(g) C at 200 °C, the Kp = 4.0 x 10-2. Find Ke Sample Solution Given T = 200 °C = 273 + 200 = 473 K Kp = 4.0 x 10-2 ∆ n = 1 - 3 = -2 Effect of a Temperature Change on the Equilibrium Constant The change in temperature will affect the value of the equilibrium constant in the following ways: For an exothermic reaction: • An increase in temperature will lower the Ke. For an endothermic reaction: • An increase in temperature will raise the Ke....
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This note was uploaded on 07/06/2011 for the course SCH 4U taught by Professor White during the Spring '10 term at Beacon FL.

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Formation of Precipitate and Ke to Kp - The relationship...

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