KEY W7 - Free Eergy S14

# KEY W7 - Free Eergy S14 - N am es_3 0 ptS total CHEM 104...

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Unformatted text preview: N am es:____3 0 ptS total___ CHEM 104 Workshop 7: Free Energy Recall that AG = AGO + RTan and that at equilibrium AG = 0 and Q 9 K. This allows us to relate free energy to K as follows: AGO = -RT In K We also know that AG0 = AH0 - TASO, where, over moderate temperature ranges [ranges less than 100 K degrees) both A50 and AH0 can be considered approximately temperature independent. In todays workshop you will use this relationship to extract values of enthalpy and entropy using temperature dependent measurements of the equilibrium constant. Lets tr and solve some roblems: 20 oints 1. By measuring the pH at various temperatures, the Kb for Temperature (K) n NH3[aq) was found to be temperature dependent, yielding - the values in the table. (15 pts) 283 1.34 X 10'5 a] By plotting the available information appropriately, 293 1.42 X 10—5 obtain estimates for A80 and AH0 for the reaction: NH3(aq) + H20 2 NH4+(aq) + 0H-(aq) 303 1.50 X 10-5 HINT 1: To begin this problem you must find an equation that relates Kb to A50 and AH0 [Think about combing the equations above). Once you have done this, you will need to put the equation in the linear form ofy = mx + b, and plot it. HINT 2: choosing your variables can be complicated. Clearly K will change when you change T. So make T your independent variable [x], and then an would be your dependent variable [y]. A general approach is to isolate your 'Y' on the left, and then look at what multiplies your 'X’ [that would be the slope, m) and what is left over [that would be your y-intercept, b). Make sure you understand which two elements are your variables [i.e., your x and y) and which elements are constants. Use the attached graph paper to make this plot. RT In K: = AHO - TASo In K = -[AH0/R] [1 /T] + ASU/R (2 pts for equation) l‘fTemp 1“ Kb 0.0035 -11.25 If the entropy and enthalpy are approximately temperature independent over the range of temperatures plotted, you should 0-0034 '11-16 obtain a straight line. The slope will equal -AH°/R, and the y- intercept will equal ASO/R. 0.0033 -11.107 See graph below and data above to solve for: Slope = -484 = -AH9/R 9 AH0 = 4.02 kl/mol (4 pts) y-intercept = -9.51=ASQ/R 9 AS0 = -79.1]/mol-K (4 pts) Names:___30 pts total -11.1 -11.12 -11.14 -11.16 —11.18 —11.2 -11.22 -11.24 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035 0.00355 1/T(K-1) b] Are your results consistent with what you would predict for the temperature dependence ofthe equilibrium constant based on Le Chatelier's principle? [5 pts] Yes, because reaction is endothermic, and K increases with increasing temperature. EXTRA CREDIT: [5 ptsl Ifthe equilibrium position ofa chemical reaction shifts towards products as the temperature increases, which ofthe following MUST be true? Circle all those that MUST be true and explain why. a] AH°<O b] AH°>0 If K increases with increasing temperature, then the reaction must be endothermic c] AS°<0 d] AS°>O e] AG°<0 f]AG°>U ...
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