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Unformatted text preview: CH201 SI: CHAPTER 4 PRACTICE TEST SOLUTIONS  For the following reaction at 298K:  2NO2(g)   N2O4(g)  1. Determine the standard enthalpy and standard entropy for the reaction from  tabulated values.  ΔH0  = ΔH 0f (N2O4(g)) – 2* ΔH 0f (NO2(g)) = 9.66 kJ – 2(33.84 kJ) = ­58.02 kJ  ΔS0 = S0(N2O4(g)) – 2* S0(NO2(g)) = 304.3 J*K­1 – 2(240.45 J*K­1) = ­176.6 J*K­1  ΔH0 =  ‐58.02 kJ  ΔS 0 = ‐176.6 J*K‐1  2. Use standard enthalpy of reaction and standard entropy of reaction to determine  the standard free energy of the reaction.  Find Keq for the reaction.  ΔG0 = ΔH0 ­ TΔS0  = ­58.02 kJ – 298 K(­0.1766 kJ*K­1) =   ­5.39 kJ  Keq = e^(­ΔG0/RT) = e­(­5.39 kJ)/(0.0083 kJ*mol­1*K­1 * 298K) = 8.84  ΔG 0 = ‐5.39 kJ  Keq = 8.84  3. Circle All correct statements about this reaction:        a. The reaction is exothermic.    The sign of ΔH0 is negative. Thus, the reaction is exothermic  b. The free energy of the products is greater than the reactants.  ΔG0 is defined as the reaction above with initial amounts 1.0 atm NO2  and 1.0 atm N2O4.  This effectively takes concentration out of  consideration allowing direct comparison of the free energies between  products and reactants.  In this case ΔG0 is negative, so the free energy of  the products is lower than the reactants.    c. As temperature increases the reaction becomes less extensive.  This is assuming ΔH0 and ΔS0 are independent of temperature.  From the  equation ΔG0 = ΔH0 ­ TΔS0 it should be apparent that ΔG0 increases with  increasing temperature (i.e. becomes more positive).  The more positive  ΔG0, the less extensive the reaction.    d. As temperature decreases Keq decreases.  Using the same reasoning as c, as temperature decreases ΔG0 becomes  more negative.  As ΔG0 becomes more negative, Keq increases.            4. Determine Q for this reaction if PNO2 = 0.24 atm and PN2O4 = 1.73 atm.  Q = (PN2O4/1 atm)/(PNO2/1 atm)2 = (1.73 atm/1 atm)/(0.24 atm/1 atm)2  = 30.03    Q = 30.03  5. Determine ΔG for the reaction from #4.  ΔG = RT ln(Q/K) = (0.0083 kJ*mol­1*K­1)(298 K) ln (30.03/8.84) = 3.02 kJ    ΔG = 3.02 kJ  6. Circle the correct answer.  This reaction is spontaneous to the:    LEFT    /  RIGHT   /  NEITHER  e. At 298K equilibrium favors products.    See b.  A negative ΔG0 tells us equilibrium favors products.  f. All of the above statements are incorrect.    I’m not this cruel.  Note that not only does the sign of ΔG tell you which direction the reaction is  spontaneous, but the relationship between Q and K does as well.    7. See previously emailed practice test.  a. Note the reaction must be spontaneous to the left.  This allows c to be  eliminated.  The difference between a and b is where equilibrium lays.  From  ΔG0 it is obvious equilibrium lays towards the products which is consistent with  a.  Realize the free energy of the products in a is lower than the free energy of  the reactants as discussed in question 3.  8. Determine the equilibrium pressures of NO2 and N2O4 at equilibrium for the  reaction from #4. (This type of problem will be covered in Chapter 5, but I’ve  included it here for completeness.  After we cover Chapter 5 material look back over  this problem and see how it fits into the ‘scheme’ of this worksheet)        I will go over this problem after we discuss chapter 5.  For now you should be able to  determine which direction the reaction will proceed (i.e. whether products or  reactants are consumed and whether products or reactants are produce to reach  equilibrium).    PNO2 = __________ atm  PN2O4 = __________ atm  ...
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