hw3 soln - CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL...

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17-1 CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS FOLLOW-UP PROBLEMS 17.1 Plan: First, balance the equations and then use the coefficients to write the reaction quotient as outlined in Equation 17.4. Solution: a) Balanced equation: 4 NH 3 ( g ) + 5 O 2 ( g ) ' 4 NO( g ) + 6 H 2 O( g ) Reaction quotient: Q c = [] [ ] 46 2 4 5 32 N OH O NH O   b) Balanced equation: 3 NO( g ) ' N 2 O( g ) + NO 2 ( g ) Reaction quotient: Q c = [ ][ ] 22 3 NO NO NO Check: Are products in numerator and reactants in denominator? Are all reactants and products expressed as concentrations by using brackets? Are all exponents equal to the coefficient of the reactant or product in the balanced equation? 17.2 Plan: Add the individual steps to find the overall equation. Write the reaction quotient for each step and the overall equation. Multiply the reaction quotients for each step and cancel terms to obtain the overall reaction quotient. Solution: (1) Br 2 ( g ) ' 2 Br( g ) (2) Br( g ) + H 2 ( g ) ' HBr( g ) + H( g ) (3) H( g ) + Br( g ) ' HBr( g ) Br 2 ( g ) + 2 Br ( g ) + H 2 ( g ) + H ( g ) ' 2 Br ( g ) + 2 HBr( g ) + H ( g ) Canceling the reactants leaves the overall equation as Br 2 ( g ) + H 2 ( g ) ' 2 HBr( g ). Write the reaction quotients for each step: Q c1 = 2 2 Br Br Q c2 = [ ][ ] [ ] 2 HBr H Br H Q c3 = [ ] HBr HB r and for the overall equation: Q c = 2 HBr r Multiplying the individual Q c ’s and canceling terms gives Q c1 Q c2 Q c3 = 2 2 Br Br [ ][ ] 2 HBr H Br H [ ] HBr r = 2 HBr r The product of the multiplication of the three individual reaction quotients equals the overall reaction quotient. 17.3 Plan: When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. Solution: a) All coefficients have been multiplied by the factor 1/2 so the equilibrium constant should be raised to the 1/2 power (which is the square root). For reaction a) K c = (7.6 x 10 8 ) ½ = 2.7568 x 10 4 = 2.8x10 4
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17-2 b) The reaction has been reversed so the new K c is the inverse of the original equilibrium constant. The reaction has also been multiplied by a factor of 2/3, so the inverse of the original K c must be raised to the 2/3 power. For reaction b) K c = 2 3 8 1 7.6 x 10    = 1.20076 x 10 -6 = 1.2 x 10 -6 Check: The order of magnitude of each answer can be estimated by examining the exponents: for part a) (10 8 ) 1/2 = 10 4 and for part b) rounding 7.6 x 10 8 to 10 9 gives 1/10 9 = 10 -9 and (10 -9 ) 2/3 = 10 -6 . These estimates are within the same order of magnitude as the answers under solution. 17.4 Plan: K p and K c for a reaction are related through the ideal gas equation as shown in K P = K c (RT) n . Find n gas , the change in the number of moles of gas between reactants and products (calculated as products minus reactants). Then, use the given K c to solve for K p . Solution: The total number of product moles of gas is 1 and the total number of reactant moles of gas is 2. n = 1 - 2 = -1. K P = K c (RT) n K P = 1.67[(0.08206) (500.)] - K P = 0.040701925 = 4.07 x 10 -2 Check: Although equilibrium constants do not have units, it is useful to check this type of calculation by doing unit analysis. For this purpose add units for the equilibrium constants and substitute M for mol/L in gas constant.
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hw3 soln - CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL...

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