S11-HW3-CH14-15

S11-HW3-CH14-15 - HOMEWORK WEEK 3 14.8 The reaction...

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HOMEWORK WEEK 3 CHAPTERS 14, 15 14.8 The reaction quotient, Q c , is an expression that has the same form as the equilibrium- constant expression but whose concentrations are not necessarily equilibrium concentrations. It is useful in determining whether a reaction mixture is at equilibrium or, if not, what direction the reaction will go as it approaches equilibrium. As a reaction occurs, Q changes and continues to do so until equilibrium is reached, at which point Q = K. This is illustrated on our Excel spreadsheets. 14.62 Calculate Q, the reaction quotient, and compare it to the equilibrium constant (K c = 3.59). If Q is larger, the reaction will go to the left; smaller, the reaction will go to the right; equal, the reaction is at equilibrium. In all cases, Q is found by combining these terms: Q = 2 2 4 4 2 2 S] H [ ] [CH ] [H ] CS [ a. Q = 2 4 2) (1.26)(1.3 ) 12 . 1 )( 43 . 1 ( = 1.02 49 = 1.02 Q < K c . The reaction should go to the right; Q rises until it equals K c b. Q = 2 4 2) (1.25)(1.5 ) 73 . 1 )( 15 . 1 ( = 3.56 6 = 3.57 Q = K c . The reaction is at equilibrium. c. Q = 2 4 1) (1.20)(1.3 ) 85 . 1 )( 15 . 1 ( = 6.54 1 = 6.54 Q > K c . The reaction should go to the left. d. Q = 2 4 3) (1.56)(1.4 ) 91 . 1 )( 23 . 1 ( = 5.13 14 = 5.13 Q > K c . The reaction should go to the left; Q decreases until it equals K c 14.74 a. Forward direction (Left-to-right): Cl 2 is on the left side of the equation. Increasing Cl 2 causes reaction to occur to remove some of this added Cl 2 . Using the reaction quotient, Q: The Q expression is: Q c = 5 3 2 [PCl ] [PCl ] x [Cl ] . At equilibrium, Q c = K c . (a) If Cl 2 is added to the equilibrium mixture, this increases the denominator of the Q c expression, which decreases the value of Q c . Q c is now smaller than K c (which is a constant and does not change!!). As Q c < K c , reaction occurs left-to-right to restore equilibrium.

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17.46 Calculate Q c , the reaction quotient of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Q c with K c to determine whether precipitation has occurred. a) Sr 2+ (aq) + CO 3 2- (aq) SrCO 3 (s), K c = 1.08 x 10 9 Q c = 2+ 2- 3 1 1 = (0.012) (0.0015) [Sr ] [CO ] = 5.6.x 10 4 Here, Q c < K c , implying that reaction occurs left-to-right. This corresponds to formation of a precipitate. b)
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This note was uploaded on 02/20/2011 for the course CHEM 110 taught by Professor Dr.huston during the Spring '06 term at Pittsburgh.

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S11-HW3-CH14-15 - HOMEWORK WEEK 3 14.8 The reaction...

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