Final Exam Review - Chem 001B Cumulative Worksheet Key PDF

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Jack Eichler (Spring 2018)...

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Chem001B Cumulative Worksheet Name: Section/Time: SHOW ALL WORK TO RECEIVE CREDIT!!! Acid–base reactions can have a strong environmental impact. For example, a dramatic increase in the acidity of rain and snow over the past 150 years is dissolving marble and limestone surfaces, accelerating the corrosion of metal objects, and decreasing the pH of natural waters. Acid rain is caused by nitrogen oxides and sulfur dioxide produced by both natural processes and the combustion of fossil fuels. Eventually, these oxides react with oxygen and water to give nitric acid and sulfuric acid. SO2(g)+H2O(l)→H2SO3(aq) H2SO3(aq)+ 1/2 O2(g)→ H2SO4(aq) 1. Write the overall reaction for the above steps. SO2(g) + H2O(l) + ½O2H2SO4(aq) 2. Draw the Lewis dot structure of each of the compounds in the overall reaction. Draw the net dipole if any. SO2 H2O O2 H2SO4

Net dipole towards oxygen

towards oxygen no net dipole

toward OH groups

3. Determine the strongest intermolecular force for each of the reactants in the overall reaction. Which intermolecular force will be the dominant force when all of the reactants are mixed in a droplet of water?  SO2 = dipole-dipole  H2O = hydrogen bonding  O2 = induced dipole-induced dipole  (hydrogen bonding is the most dominant force for the reaction to occur) 4. The concentration of SO2 (MW = 64 g/mol) in air for populated/industrious regions is above 2 mg per 1000 mL (aka > 2 ppm). From this information determine the atmospheric molarity of SO2 and its partial pressure in the lower atmosphere at 10o C for an altitude of ~3000 m.  Calculate the moles of SO2 first Mol SO2= (2mg)x(1g/1000mg)x(1mol/64g)=3.125E-5 mol SO2 Molarity=(3.125E-5mol)/(1L)=3.125E-5M  Assume a 1L atmospheric solution for SO2 (PSO2)V=nSO2RT (PSO2)(1L)=(3.125E-5mol)(0.08206atm*L/mol*K)(273+10K) PSO2=7.26E-4atm

5. Free SO2 can dissolve into water droplets of vapor and clouds, eventually converting to highly acidic H2SO4. Given KH = 1.5 mol/L*atm for SO2 at the indicated temperature, use the partial pressure you obtained in problem 4 to calculate the concentration of dissolved SO2 in a water droplet. Assuming 100% of the dissolved SO2 is then converted to H2SO4 (MW =98.08 g/mol), what is the molality of this strong acid in these droplets?  Henry’s Law= Cg=kHPg Cg=1.5mol/L*atm*7.26E-4 atm Cg=0.001089E-4 M  For molality calculation: assume you have 1L droplet of water (that’s a large droplet) Density of water = 1.00g/mL 1L=1000mL=1000mL*1.00g/mL=1000g of solution  Molarity of SO2= 7.25E-4 mol/L=Molarity of H2SO4 Solve for mol and convert to grams of H2SO4 (7.25E-4mol/L)*(1L soln)=7.25E-4mol*98.08g/mol=0.071 g H2SO4  Solve for kg of solvent= 1000g-0.071g H2SO4=999.9262236 g water  Molality=mol solute/kg solvent=(7.5221E-4mol SO2/0.9999262236)=7.5226E-4 m 6. Let’s have a look at the reactant concentration dependence for the formation of sulfuric acid. Using the following data, calculate the order of the reaction and the rate constant. Experiment

SO2 (M)

O2 (M)

H2O (M)

Initial Rate of Reaction (M/s)

1 0.001 0.004 0.001 4.0E-13 2 0.002 0.004 0.001 8.0E-13 3 0.003 0.002 0.002 3.0E-13 4 0.003 0.001 0.002 7.5E-14 5 0.004 0.003 0.002 9.0E-13 6 0.004 0.003 0.004 9.0E-13 Rate=k[SO2]a[O2]b[H2O]c Compare the following experiments: T1 to T2, T3 to T4 and T5 to T6 T2=(0.008)=k(0.002)a(0.004)b(0.001)c T1=(0.004)=k(0.001)a(0.004)b(0.001)c =2=2aa=1 Repeat: b=2, c=0 Rate Law= Rate=k[SO2][O2]2 order of reaction is 3 k= 2.5E-5 7. Monitoring the rate constant as a function of temperature leads to a collision frequency of 0.115. Using the k value obtained from above, calculate the energy of activation for the reaction at 10o C.  Arrhenius equation= ln(k)=-Ea/R(1/T) +ln(A) where R=8.314 (J/mol*K) -RT(ln(K)-ln(A))=Ea -(8.314)*(273+10)(ln(-2.5E5)-ln(0.115))=Ea -(8.314)*(283)(-8.4338)=Ea 19843.56754J/mol=Ea=19.8kJ/mol

8. Metal corrosion by aqueous acid is a favorable process under many normal conditions and one of the many problems caused by acid rain. Use the Following reaction and tabulated values to calculate H and S for the following reaction between metallic iron and free acid generated by species such as H2SO4. What is the overall favorability of the reaction (G) assuming STP? Fe (s) + 2H(aq) +  Fe(aq)2+ + H2(g) Species fo (kJ/mol) So (J/mol*K)   Fe   H2 2+ Fe -89.1 -137.7 H+ 0 0  Calculate fo from the table fo=summation of products-summation of reactants fo= [1(0)+1(-89.1)]-[1(0)+2(0)]= -89.1 kJ/mol  Calculate So from the table So= summation of products-summation of reactants So=[1(130.68)+1(-137.7)]-[1(27.28)+2(0)]= -34.3 J/mol  Calculate G=H-TS G=(-89100)-(298(-34.3))=-78878.6J/mol=-78.9kJ/mol Spontaneous reaction at STP...