PracticeExam2

PracticeExam2 - Chem 5 Winter, 2005 Practice Second Exam...

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Chem 5 Winter, 2005 Practice Second Exam (the actual second exam last year) Page 1 of 6 1. (5 points each) Here are a few quick problems to get you going. (a) What is the pH of a saturated solution of Ca(OH) 2 , for which K sp = 1.3 × 10 –6 ? Let the equilibrium concentration of Ca 2+ , [Ca 2+ ], equal x . Then [OH ] = 2 x , and the equilibrium constant expression is K sp = x (2 x ) 2 = 4 x 3 . Solving for x yields x = 6.87 × 10 –3 M, so [OH ] = 2 x = 1.38 × 10 –2 M and pOH = –log([OH ]) = 1.86 so pH = 14.00 – pOH = 12.14. (b) The reaction enthalpy for the formation of one mole of iridium sulfide, Ir 2 S 3 ( s ), from Ir( s ) and S 2 ( g ) is –400.68 kJ mol –1 , and the standard molar enthalpy of formation of S 2 ( g ) is 128.37 kJ mol –1 . What is the standard molar enthalpy of formation of Ir 2 S 3 ( s )? The synthesis reaction is 2 Ir( s ) + 3 2 S 2 ( g ) Ir 2 S 3 ( s ) This would be the formation reaction for Ir 2 S 3 , (iridium is a metallic element) except the standard state for sulfur is S( s ) instead of S 2 ( g ), as the fact that the S 2 enthalpy of formation is not zero suggests! Nevertheless, we can write the following expression for the enthalpy change for this reaction: H r o = –400.68 kJ mol –1 = H f o (Ir 2 S 3 ) – 2 H f o (Ir) – 3 2 H f o (S 2 ) = H f o (Ir 2 S 3 ) – 2 × 0 – 3 2 × (128.37 kJ mol –1 ) We solve this and find H f o (Ir 2 S 3 ) = –208.13 kJ mol –1 . (c) How many moles of Ag 2 CrO 4 ( K sp = 9.0 × 10 –12 ) will dissolve in 1.00 L of a 0.09 M solution of Na 2 CrO 4 ? We write the equilibrium constant expression with chromate ion at [CrO 4 2– ] = 0.09 M: K sp = 9.0 × 10 –12 = [Ag + ] 2 [CrO 4 2– ] = [Ag + ] 2 (0.09) This tells us [Ag + ] = 10 –5 M, and because every mole of Ag 2 CrO 4 that dissolves produces two moles of Ag + in solution, the molar solubility of the solid is 5 × 10 –6 mol L –1 .
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Chem 5 Winter, 2005 Practice Second Exam (the actual second exam last year) Page 2 of 6 2. (4 + 6 + 6 + 4 points) A sample of a fixed amount of an ideal gas is subjected to a series of state changes as shown in the P , V diagram below. The gas starts in state at a temperature of 200 K, a pressure of 4 atm, and a volume of 1 L. The dashed lines in the diagram show how the external pressure varied as the gas was taken to a series of three intermediate equilibrium states, through , before ending at the final equilibrium state . The P , V coordinates of each state is indicated by a black dot. 5 4 3 2 1 0 012345 P /atm V /L (a) What are the temperatures of the gas at each intermediate state?
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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PracticeExam2 - Chem 5 Winter, 2005 Practice Second Exam...

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