Thermodynamics - Principles of Chemistry - Exam, Exams of Chemistry

Download Thermodynamics - Principles of Chemistry - Exam and more Exams Chemistry in PDF only on Docsity! Exam Thermodynamics 1st, 2nd, and 3rd Laws Enthalpy, Entropy, and Gibbs Energy Heats of Formation Hess’ Law Spontaniety Gibbs Equation Electrochemical Cells Balancing Redox Reactions Nernst Equation Batteries Electroplating and Hydrolysis Docsity.com Thermodynamics 1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of water (250 mL) initially at 25°C. When the system reached thermal equilibrium the final temperature of the water and the block was 61°C. What was the initial temperature of the block of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O) = 18 g/mol, density of water = 1 g/ml 2) Given the following information calculate the heat of formation of C2H4. C2H4 + 3 O2 → 2 CO2 + 2 H2O ΔH° = -414 kJ/mol C + O2 → CO2 ΔH° = -393.5 kJ/mol H2 + ½ O2 → H2O ΔH° = -241.8 kJ/mol 3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH? 4) Given a block of ice at 0°C how much of the ice will melt if you put a 100 gram block of aluminum on it that is initially at 100°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol 5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess oxygen. The balanced combustion reaction was as follows; C4H10O + 6 O2 -----> 4 CO2 + 5 H2O The heat released by this reaction heated 2500 mL of water 8.6°C. What is the heat of formation of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, ΔH°f(CO2) = -393.5 kJ/mol, ΔH°f(H2O) = -241.8 kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol. 6) If a 25 gram block of copper at 45°C is added to 50 grams of liquid ammonia at -60°C, calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, ΔHvap = 23.4 kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol 7) You have 50 grams of ice at -2.5°C sitting in a dixie cup. If you heat 5 nickels and throw them into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 grams and has a heat capacity of 24.3 J/mol-K. Assume that nickels are pure nickel with an atomic mass of 57.81 g/mol. 8) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The enthalpy for this reaction is - 210 kJ/mol. Docsity.com 17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the cell and lable the anode and cathode. You may use a carbon electrode for the I- solution. Fe3+ + e- →> Fe2+ +0.771 V I2 → 2 I- +0.535 V 18) What is the value of the half reaction, Cu2+ + e- → Cu+ given that, Cu2+ + 2e- → Cu° ° = +0.3402 Cu+ + e- → Cu° ° = +0.522 19) Given the following half-reactions draw an electrochemical cell that would work. Calculate the voltage of the cell and label the anode and cathode, tell which electrode is positive and which is negative, and where the oxidation and reduction reactions are occurring. In addition indicate the direction of electron flow, and the concentration of any ionic species in solution. Co3+ + e- → Co2+ +1.842 V Pb2+ + 2e- → Pb -0.126 V 20) Given the following half-reactions, properly set up a WORKING electrochemical cell. Ag + I- → AgI + e- ° = 0.1519 V PbI2 + 2e-→ Pb + 2 I- ° = 0.3580 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate what the electrodes are made of. 20b) Draw the cell diagram for the above cell. 20c) What is the Keq for the above cell? 20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is reasonable or not. Docsity.com 21) Balance the following half-reactions in a BASE and then use these reactions to set up a WORKING electrochemical cell. Bi(s) → Bi2O3(s) 0.460 Volts Hg2O(s) → Hg(l) -0.123 Volts Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 21b) Draw the cell diagram for the above cell. 21c) What is the Keq for the above cell? 22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10 amps? MW (Au) = 197.8 23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a constant current of 25 amps? MW (Pd) = 106.4 Docsity.com Thermodynamics 1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of water (250 mL) initially at 25°C. When the system reached thermal equilibrium the final temperature of the water and the block was 61°C. What was the initial temperature of the block of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O) = 18 g/mol, density of water = 1 g/ml -nCpTCu = nCpTH2O Ti = 1036.3°C 2) Given the following information calculate the heat of formation of C2H4. C2H4 + 3 O2 → 2 CO2 + 2 H2O Δ H° = -414 kJ/mol C + O2 → CO2 Δ H° = -393.5 kJ/mol H2 + ½ O2 → H2O Δ H° = -241.8 kJ/mol 2 CO2 + 2 H2O → C2H4 + 3 O2 ΔH° = 414 kJ/mol 2 (C + O2 → CO2) ΔH° = -393.5 kJ/mol x2 2 (H2 + ½ O2 → H2O) ΔH° = -241.8 kJ/mol x2 2 C + 2 H2 → C2H4 ΔH° = 414 kJ/mol + 2 (-393.5 kJ/mol )+ 2(-241.8 kJ/mol) ΔH° = -856.6 kJ/mol 3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH? C2H5OH + 3 O2 → 2 CO2 + 3 H2O ΔH° = ΔH°prod - ΔH°react -1237.7 kJ/mol= [2(-393.5kJ/mol) + 3(-241.8kJ/mol)] – [ΔH°C2H5OH + 3(0kJ/mol)] ΔH°C2H5OH = -274.7 kJ/mol    C25-C61 molK J2.57 g/mol18 250gTi-C61 molK J24.5 g/mol63.55 100g              Docsity.com 9) Calculate the ΔG for the making of 6M NaOH from solid NaOH. Saturated solutions of NaOH have a concentration of 19.25 M. There are two processes going on, 1. Solid NaOH ↔ Sat’d NaOH 2. Sat’d NaOH → 6 M NaOH The first process is an equilibrium so G = 0 for the first process. For the second process, since all solutions have a standard state of 1M, the standard state for Sat’d NaOH and 6 M NaOH are both 1 M and the G° = 0 since there is no difference between the G°’s for these two solutions. So, G1 = 0 G2 = G° + RT ln Q G2 = 0 + (8.314 J/molK)(298K) ln (6 M/ 19.25M) = -2888.2 J/mol G1+ G2 = 0 + (- 2888.2 J/mol) = - 2888.2 J/mol 10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign? Work = PV A chemist thinks of himself as being inside the system pushing out (like expanding a balloon). To a chemist, pushing out (expanding the balloon) is negative work. Negative work is work that you must do as opposed positive work that is done for you. In mathematics, a change in volume is always the Vfinal – Vinitial. In the case of an expanding balloon, the final volume will be larger than the initial volume so Vfinal – Vinitial will produce a positive value. Since PV = negative for an expanding balloon, and both P and V are positive, we must add a negative sign to the work expression to make this work overall negative. So, chemists define pressure-volume work as, Work = - PV (Chemist) It should be noted that engineers see the world opposite from chemists. Engineers view themselves as being outside of the system pushing in. Work for an engineer is compressing the balloon, making it smaller. Since V = Vfinal – V initial, and the Vfinal is smaller than Vinitial, the V is negative. So, for an engineer, Work = PV (Engineer) This value is still negative, like the chemist, but we see the world differently from an engineer. Docsity.com 11) Please give me a definition of a heat capacity. It is the amount of energy needed to raise the temperature of one mole of substance 1°C. 12) What is the ΔG° for the following reaction at 25°C? BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s) ΔH° S° BaBr2 -186.47 kJ/mol 42.0 J/mol K CuSO4 -201.51 kJ/mol 19.5 J/mol K BaSO4 -345.57 kJ/mol 7.0 J/mol K CuBr2 -25.1 kJ/mol 21.9 J/mol K ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react ΔH° = ΔH°prod - ΔH°react ΔH° = [(-345.57 kJ/mol) + (-25.1 kJ/mol)] – [(-186.47 kJ/mol) + (-201.51 kj/mol)] ΔH° = 17.31 kJ/mol ΔS° = S°prod - S°react ΔS° = [(7.0 J/molK) + (21.9 J/molK)] – [(42.0 J/molK) + (21.9 J/molK)] ΔS° = -32.6 kJ/mol G° = H° - TS° G° = 17,310 J/mol + (298 K) (-32.6 J/molK) G° = 27,024.8 J/mol Circle all that apply about this reaction, spontaneous not spontaneous exothermic endothermic entropy increases entropy decreases the reaction occurs quickly the reaction occurs slowly What is the equilibrium constant for this reaction? G° = - RT ln Keq 5-8)(8.314)(29 )8.024,72( RT ΔG x10832.1eeKeq   Docsity.com 13) Given the following set of thermodynamic data calculate the ΔG°, ΔH°, ΔS°, and Keq for the following reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k) CoCl2(s) ↔ Co2+ + 2 Cl- ΔH° S° CoCl2 -325.5 106.3 Co2+ -67.36 -155.2 Cl- -167.4 55.1 ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react ΔH° = ΔH°prod - ΔH°react ΔH° = [(-67.36 kJ/mol) + 2(-167.4 kJ/mol)] – [(-325.5 kj/mol)] ΔH° = -76.66 kJ/mol (exothermic) ΔS° = S°prod - S°react ΔS° = [(-155.2 J/molK) + 2(55.1 J/molK)] – [(106.3 J/molK)] ΔS° = -151.3 J/molK (entropy decreases) G° = H° - TS° G° = -76,660 J/mol - (298 K) (-151.3 J/molK) G° = -31,572.6 J/mol (reaction is spontaneous as written) G° = - RT ln Keq 58)(8.314)(29 31,572.6)( RT ΔG 3.423x10eeKeq   Docsity.com 16) Consider the following electrochemical reaction, Au → Au3+ + 3e- AuCl4- + 3e- → Au + 4 Cl- AuCl4- → Au3+ + 4 Cl- How would the voltage change if you, a) Enlarge the Au electrodes? Inc. N.C. Dec. b) Add KCl(s)? Inc. N.C. Dec. c) Add AgNO3? Inc. N.C. Dec. d) Add water to the reduction cell? Inc. N.C. Dec. e) Add NaNO3(s) Inc. N.C. Dec. f) Cool the reaction vessel? Inc. N.C. Dec. Note: This is NOT a Ksp problem since AuCl4- is not a solid (solids are not charged) so we will leave the reaction as is.  = ° - RT/n ln [Au3+][Cl-]4/[ AuCl4-] a) No change. Changing the size of the electrode makes no difference. b) Adding Cl- makes the natural log large which decreases the voltage c) The Ag+ will react with the Cl- to make AgCl. This removes Cl-, making the natural log smaller (more negative). So you will be subtracting a smaller number so the voltage will increase. d) Diluting the reduction cell lowers the concentration of both AuCl4- and Cl- but the Cl- is raised to the 4th power so it reduces faster. Reducing Cl- makes the natural log smaller (more negative so you are subtracting a smaller (more negative) number so voltage increases. e) No change. f) If T is smaller, then you are subtracting a smaller number so voltage increases. Docsity.com 17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the cell and label the anode and cathode. You may use a carbon electrode for the I- solution. Fe3+ + e- → Fe2+ +0.771 eV 2 I- → I2 + 2 e- -0.535 eV 2 Fe3+ + 2 I- → 2 Fe2+ + I2 0.236 eV 18) What is the value of the half reaction, Cu2+ + e- ----> Cu+ given that, Cu2+ + 2e- → Cu ° = +0.3402 Cu+ + e- → Cu ° = +0.522 Cu2+ + 2e- → Cu ° = +0.3402 Cu → Cu+ + e- ° = -0.522 Cu2+ + e- → Cu+ ° = 2(0.3402) + 1(-0.522) = 0.1584 volts 1 C Pt 1M KI 1M FeCl 1M FeCl 3 2 e- Anode Negative Oxidation Cathode Positive Reduction I2 Docsity.com 19) Given the following half-reactions draw an electrochemical cell that would work. Calculate the voltage of the cell and label the anode and cathode, tell which electrode is positive and which is negative, and where the oxidation and reduction reactions are occurring. In addition indicate the direction of electron flow, and the concentration of any ionic species in solution. Co3+ + e- → Co2+ Δ° = + 1.842 eV Pb2+ + 2e- → Pb Δ° = - 0.126 eV Co3+ + e- → Co2+ Δ° = + 1.842 eV Pb → Pb2+ + 2e- Δ° = + 0.126 eV 3 Pb + 2 Co3+ → 3 Pb2+ + 2 Co Δ° = 1.968 V Pb Pt 1M Pb(NO )3 2 1M CoCl 1M CoCl 3 2 e- Anode Negative Oxidation Cathode Positive Reduction Docsity.com 22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10 amps? MW (Au) = 197.8 31.1 g/197.8 g/mol = 0.1572 mol Au has a 3+ charge, so multiply by 3. 3 mole e-/mol Au x 0.15723 mol Au = 0.4717 mol e- 0.4717 mol e- x 96,487 C/mol e- = 45,511.8 C Amp x sec = Coulombs (C) (10 amps)(sec) = 45,511.9 C sec = 4,551.18 sec => 75.85 minutes 23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a constant current of 25 amps? MW (Pd) = 106.4 453.6 g/106.4 g/mol = 4.263 mol Pd, but Pd has a 4+ charge, so multiply by 4. 4 mole e-/mol Pd x 4.263 mol Pd = 17.053 mol e- 17.053 mol e- x 96,487 C/mol e- = 1,645,357 C Amp x sec = Coulombs (C) (25 amps)(sec) = 1,645,357 C sec = 65,814 sec => 18.28 hours Docsity.com Chemistry 121 Name____________________ Third Exam May 15, 2007 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 ( 18) 2 (40) 3(12) 4(15) 5(15) Total R = 8.314 J/mol-K  = 96486 C/mol e- C = amp x sec  =  - 0.0592/n log Q ΔG = ΔG + RT ln Q ΔG= -RTlnKeq ΔG = -n PV = nRT R = 0.08205 L-atm/mol-K 1) Balance each of the following oxidation-reduction half-reactions; IO3- —> I3- (in acid) Mn(OH)2 —> Mn2O3 (in base) Docsity.com 2) Given the following set of half-reactions, set up a WORKING electrochemical cell. Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl-  = 0.27 V 2 H+ + 2 e- —> H2  = 0.000 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2b) Draw the cell diagram for the above cell. 2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M? Docsity.com 2) Given the following set of half-reactions, set up a WORKING electrochemical cell. Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl-  = 0.27 V 2 H+ + 2 e- —> H2  = 0.000 V Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2b) Draw the cell diagram for the above cell. (Note: The solution in both cells is HCl so no salt bridge is required). Pt/H2 (1atm) / 1M HCl / Hg2Cl2(s) / Hg(l)/ Pt 2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M? The overall reaction is, Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl-  = 0.27 V H2 —> 2 H+ + 2 e- = 0.000 V Hg2Cl2(s) + H2 –> 2 Hg + 2 H+ + 2 Cl- = 0.27 V So the voltage is,  = 0.27 - 0.0592/2 log [H+]2 [Cl-]2 = 0.27 - 0.0591/2 log [0.10]2 [0.10]2 = 0.3882 Volts H2 1atm 1M HCl Pt Hg Cl2 2 1M HCl Pt e- Anode Negative Oxidation Cathode Positive Reduction Docsity.com 3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current (amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol 2 grams/65.41g/mol = 0.03058 mole of Zn 0.03058 mole of Zn x 2 mole e-/mol Zn = 0.06115 mole e- 0.06115 mole e- x 96,486 C/mol e- = 5900.4 Coulombs 1 C = 1 amp x sec 5900.4 C = 0.8 amps x sec 5900.4 C/0.8 amps = 7375.47 seconds 4) You are given 100 mL of water at 25°C. You add a 20 gram ice cube at 0°C and 50 mL of boiling water at 100°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol and Cp(H2O) = 75.2 J/mol K, FWT H2O = 18g/mol (These were not given in the original problem). The easiest way of doing this is by mixing the hot and cold water and then adding the ice cube. First add 100 mL of 25°C water to 50 mL of 100°C water, nCpΔT25°C = - nCpΔT100°C 100g/18g/mol (75.2 J/molK)(Tf - 25) = -50g/18g/mol(75.2 J/molK)(Tf-100) The 18g/mol and the 75.2 J/molK cancel on both sides so, 100(Tf-25) = -50(Tf-100) Tf = 50°C So now we have 150 mL of water at 50̊C and we add the 20 gram ice cube. I assume that the ice cube will melt and then waters will mix to get to some final temperature so, nΔHfus + nCpΔT0°C = - nCpΔT50°C 20g/18g/mol(6010 J/mol) + 20g/18g/mol (75.2)(Tf - 0) = - 150g/18g/mol(75.2)(Tf - 50) All the 18 g/mol cancel so, 120,200 + 1504Tf = - 11280Tf + 564000 12784Tf = 443,800 Tf = 34.71°C Docsity.com 5) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The ΔH = - 210 kJ/mol.for this reaction. ΔG = ΔG + RT ln Q But ΔG = 0 since both concentrations have the same standard state of 1M. So, ΔG = 0 + RT ln 0.03M/4.5M ΔG = -5391.42 J/mol Now, since ΔG = ΔH - TΔS, we can solve for ΔS, -5391.42 J/mol = -210,000 J/mol - (298)ΔS ΔS = (-5391.42 J/mol + 210,000 J/mol )/298 = 686.61 J/mol K Docsity.com 3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08 g/mol). What was the charge on the platinum? 4) A 20 gram block of ice initally at -10.0°C was dropped into 250 mL of water at 25°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) = 35.2 J/mol-K, FWT (H2O)= 18 g/mol Docsity.com Chemistry 121 Name__Answer Key___ Third Exam May 14, 2009 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1 (18) 2 (28) 3(10) 4(20) Total R = 8.314 J/mol-K  = 96486 C/mol e- C = amp x sec  =  - 0.0592/n log Q ΔG = ΔG + RT ln Q ΔG= -RTlnKeq ΔG = -n PV = nRT R = 0.08205 L-atm/mol-K 1a) Please balance the following half-reaction in a base. BH3 —> B2O3 3 H2O + 2 BH3 —> B2O3 + 12 H+ + 12 e- 12 OH- —> 12 OH- 12 OH- + 2 BH3 —> B2O3 + 9 H2O + 12 e- 1b) What is the voltage of the following half-cell reactions when added together? Cu2+ + e- —> Cu+  = 1.29 Cu+ + e- —> Cu  = 1.68 Cu2+ + 2 e- —> Cu  = (1.29(1) + 1.68 (1))/2 = 1.485 Volts Docsity.com Pb Pb 1M Pb(NO )3 2 1M KI PbI (s)2 e- Anode Negative Oxidation Cathode Positive Reduction 2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information below (found in the CRC), what voltage should you have obtained for your cell, and what should the Ksp have been? PbI2(s) <--> Pb2+ + 2 I- ΔH S PbI2 -174.1 kJ/mol 176.98 J/mol-K Pb2+ 1.63 kJ/mol 21.34 J/mol-K I- -55.94 kJ/mol 109.37 J/mol-K ΔH ̊ = [1.63 + 2(-55.94)] - [-174.1] = 63.85 kJ/mol Δ S = [21.34 + 2(109.37)] - [176.98] = 63.1 J/molK ΔG = 63,850 - 298(63.1) = 45,046.2 J/mol 45,046.2 J/mol = -8.314(298)ln(Ksp) Ksp = 1.27x10-8 The reaction is (Circle all that apply). Spontaneous Not Spontaneous Exothermic Endothermic Entropy increases Entropy Decreases Occurs Fast Occurs Slow 2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2. Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 2c) Draw the cell diagram for the above cell. Pb / PbI2(s) / 1 M KI // 1 M Pb(NO3)2 / Pb Docsity.com