Homework 12 (Key) First, separate into oxidation and reduction half reactions

Transcription

1 Homework 12 (Key) 1. Balance the following oxidation/reduction reactions under acidic conditions. a. MnO I - I 2 + Mn 2+ First, separate into oxidation and reduction half reactions Oxidation half reaction: I - I2 Reduction half reaction: MnO4 - Mn 2+ Next, balance all atoms except oxygen and hydrogen Oxidation half reaction: 2I - I2 Reduction half reaction: MnO4 - Mn 2+ (already done) Next, balance oxygen using water Oxidation half reaction: 2I - I2 (not necessary) Reduction half reaction: MnO4 - Mn H2O Next, balance hydrogen using H + Oxidation half reaction: 2I - I2 (not necessary) Reduction half reaction: 8H + + MnO4 - Mn H2O Next, balance charges using e - Oxidation half reaction: 2I - I2 + 2e - Reduction half reaction: 5e - + 8H + + MnO4 - Mn H2O Next, balance the number of electrons being transferred. Oxidation half reaction: 10I - 5I2 + 10e - Reduction half reaction: 10e H + + 2MnO4-2Mn H2O Combine both half reactions 10e H + + 2MnO I - 2Mn H2O + 5I2 + 10e - Cancel out anything that appears on both sides 16H + + 2MnO I - 2Mn H2O + 5I2

2 b. PbO 2 + Cl - Pb 2+ + Cl 2 First, separate into oxidation and reduction half reactions Oxidation half reaction: Cl - Cl2 Reduction half reaction: PbO2 Pb 2+ Next, balance all atoms except oxygen and hydrogen Oxidation half reaction: 2Cl - Cl2 Reduction half reaction: PbO2 Pb 2+ (already done) Next, balance oxygen using water Oxidation half reaction: 2Cl - Cl2 (not necessary) Reduction half reaction: PbO2 Pb H2O Next, balance hydrogen using H + Oxidation half reaction: 2Cl - Cl2 (not necessary) Reduction half reaction: 4H + + PbO2 Pb H2O Next, balance charges using e - Oxidation half reaction: 2Cl - Cl2 + 2e - Reduction half reaction: 2e - + 4H + + PbO2 Pb H2O Next, balance the number of electrons being transferred. Already done Combine both half reactions 2e - + 4H + + PbO2 + 2Cl - Pb H2O + Cl2 + 2e - Cancel out anything that appears on both sides 4H + + PbO2 + 2Cl - Pb H2O + Cl2

3 c. Cr 2O NO 2 - Cr 3+ + NO 3 - First, separate into oxidation and reduction half reactions Oxidation half reaction: NO2 - NO3 - Reduction half reaction: Cr2O7 2- Cr 3+ Next, balance all atoms except oxygen and hydrogen Oxidation half reaction: NO2 - NO3 - (already done) Reduction half reaction: Cr2O7 2-2Cr 3+ Next, balance oxygen using water Oxidation half reaction: H2O + NO2 - NO3 - Reduction half reaction: Cr2O7 2-2Cr H2O Next, balance hydrogen using H + Oxidation half reaction: H2O + NO2 - NO H + Reduction half reaction: 14H + + Cr2O7 2-2Cr H2O Next, balance charges using e - Oxidation half reaction: H2O + NO2 - NO H + + 2e - Reduction half reaction: 6e H + + Cr2O7 2-2Cr H2O Next, balance the number of electrons being transferred. Oxidation half reaction: 3H2O + 3NO2-3NO H + + 6e - Reduction half reaction: 6e H + + Cr2O7 2-2Cr H2O Combine both half reactions 6e H + + Cr2O H2O + 3NO2-2Cr H2O + 3NO H + + 6e - Cancel out anything that appears on both sides 8H + + Cr2O NO2-2Cr H2O + 3NO3 -

4 d. Cr 2O Fe 2+ Cr 3+ + Fe 3+ First, separate into oxidation and reduction half reactions Oxidation half reaction: Fe 2+ Fe 3+ Reduction half reaction: Cr2O7 2- Cr 3+ Next, balance all atoms except oxygen and hydrogen Oxidation half reaction: Fe 2+ Fe 3+ (already done) Reduction half reaction: Cr2O7 2-2Cr 3+ Next, balance oxygen using water Oxidation half reaction: Fe 2+ Fe 3+ (not necessary) Reduction half reaction: Cr2O7 2-2Cr H2O Next, balance hydrogen using H + Oxidation half reaction: Fe 2+ Fe 3+ (not necessary) Reduction half reaction: 14H + + Cr2O7 2-2Cr H2O Next, balance charges using e - Oxidation half reaction: Fe 2+ Fe 3+ + e - Reduction half reaction: 6e H + + Cr2O7 2-2Cr H2O Next, balance the number of electrons being transferred. Oxidation half reaction: 6Fe 2+ 6Fe e - Reduction half reaction: 6e H + + Cr2O7 2-2Cr H2O Combine both half reactions 6e H + + Cr2O Fe 2+ 2Cr H2O + 6Fe e - Cancel out anything that appears on both sides 14H + + Cr2O Fe 2+ 2Cr H2O + 6Fe 3+

5 e. Cu + NO 3 - Cu 2+ + NO 2 First, separate into oxidation and reduction half reactions Oxidation half reaction: Cu Cu 2+ Reduction half reaction: NO3 - NO2 Next, balance all atoms except oxygen and hydrogen Oxidation half reaction: Cu Cu 2+ (already done) Reduction half reaction: NO3 - NO2 (already done) Next, balance oxygen using water Oxidation half reaction: Cu Cu 2+ (not necessary) Reduction half reaction: NO3 - NO2 + H2O Next, balance hydrogen using H + Oxidation half reaction: Cu Cu 2+ (not necessary) Reduction half reaction: 2H + + NO3 - NO2 + H2O Next, balance charges using e - Oxidation half reaction: Cu Cu e - Reduction half reaction: e - + 2H + + NO3 - NO2 + H2O Next, balance the number of electrons being transferred. Oxidation half reaction: Cu Cu e - Reduction half reaction: 2e - + 4H + + 2NO3-2NO2 + 2H2O Combine both half reactions 2e - + 4H + + 2NO3 - + Cu 2NO2 + 2H2O + Cu e - Cancel out anything that appears on both sides 4H + + 2NO3 - + Cu 2NO2 + 2H2O + Cu 2+

6 2. Answer the following questions regarding the reaction a from the previous problem. a. What is being oxidized? I - c. What is the oxidizing agent? MnO4 - b. What is being reduced? Mn in MO4 - d. What is the reducing agent? I - 3. A cylinder with a movable piston contains moles of an ideal gas at C and mmhg. a. What is the volume of the container? PV = nrt V = nrt P = " ( moles) L atm % $ '(400.15) # mol K & " 1atm % $ mmhg x ' # 760 mmhg & =1.73 L b. If the temperature drops to 27.0 C and the pressure is held constant what will be the new volume? V = nrt P = " ( moles) L atm % $ '(300.15K) # mol K & " 1atm % $ mmhg x ' # 760 mmhg & =1.30 L g of XeF 4 gas is placed into an evacuated liter container at C. What is the pressure in the container? P = nrt V =! mole XeF $ 2.500g XeF 4 x 4! # & L atm $ # &(353.15K) " g XeF 4 %" mol K % = atm 3.000L ( )

7 grams of an ideal gas is placed into a 1.00 L container at 300.0K and the resulting pressure is 1.50 atm. What is the molar mass of this gas? First, determine the moles of gas n = PV RT = (1.50 atm)(1.00 L) " L atm % $ '(300.0 K) # mol K & = moles Second, determine molar mass MM = mass mole = 1.83g moles = 30.0 g mole 6. Circle the gas from each pair with the highest average velocity. a. He at 400K or CO 2 at 400K b. He at 400K or He at 500K c. 1 mole of He with (P = 1 atm, V = 1 L) or 1 mole of He with (P = 1 atm, V = 0.5 L) d. 1 mole of He with (P = 1 atm, V = 1 L) or 0.5 mole of He with (P = 1 atm, V = 1 L) 7. Which of the following gases would have an average velocity of 532 m/s at 500K? a. He b. H 2O c. CO 2 d. N 2 First, determine the molar mass v = 3RT MM v 2 = 3RT MM MM = 3RT v 2 = # J & 3% ( 500K $ mol K ' # 532 m & = kg / mol = 44.1g / mol 2 % ( $ s ' This matches the molar mass of CO2 8. Calculate the kinetic energy of mole of hydrogen gas at 450.0K. KE mole = 3 " J % RT = 1.5 $ '(450.0K) = 5612J 2 # mol K &

8 9. Calculate the average kinetic energy per molecule of hydrogen gas at 450.0K First, determine the KEmole using temperature KE mole = 3 " J % RT = 1.5 $ '(450.0K) = J 2 # mol K & Second, determine the average KE of a butane molecule Average KE = KE mole N A = J 6.022x10 23 = 9.319x10-21 J 10. Calculate the average velocity of an individual hydrogen molecule at K v = 3RT MM v = " J % 3$ '(450.0K) # mol K & 2.016x10 3 kg v = m s 11. Calculate the molecular mass (in kg) of the average molecule of methane (CH 4) g kg x mole CH g x mole CH x CH 4 molecules = x kg/molecule 12. Determine the kinetic energy of the average methane molecule traveling with a velocity of m/s. KE= 1 2 mv 2 = 1-26 (2.664 x 10 kg) 2! # m " s 2 $ & = x J % 13. Calculate the kinetic energy of moles of methane described in the previous problem x J molecule x 10 molecules x mole = 1545 J/mol

9 14. A gas mixture containing hydrogen and oxygen was placed into a ml container. The average velocity of an oxygen molecule in this mixture is 452 m/s. a. What is the temperature inside the container? v= 3RT MM v 2 = 3RT MM T= v 2 (MM) 3R ( ) 2 (32.00x10-3 kg/mol) = 452m/s J mol K b. What is the average velocity of a hydrogen molecule? = 262K v = 3RT MM v = J K mol K 2.016x10-3 kg/mol ( ) = 1.80x10 3 m/s c. How many moles of hydrogen gas are in the container if the ratio of hydrogen to oxygen is 2:1 with a pressure of 1.00 atm? n T = P T V RT = ( 1.00atm) ( 0.500L) L atm = moles (262K) mol K n H2 = 2 3 n T = 2 3 ( moles)= moles

10 grams of zinc metal are placed into a flask containing ml of 3.00 M hydrochloric acid. What volume of hydrogen gas can be generated at STP if the reaction is allowed to go to completion? The reaction between magnesium metal and hydrochloric acid is as follows Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) 8.500g Zn x mole Zn 65.39g Zn x mole H 2 mole Zn = mole H mLx V= nrt P = 3.00 mole HCl 1000mL x mole H 2 2 mole HCl = mole H 2 ( mole) L atm (273.15K) mol K = 2.914L 1.000atm grams of zinc metal are placed into a flask containing ml of 3.00 M hydrochloric acid. What volume of hydrogen gas can be generated at STP if the reaction is allowed to go to completion? 10.00g Zn x mole Zn 65.39g Zn x mole H 2 mole Zn = mole H mLx V= nrt P = 3.00 mole HCl 1000mL x mole H 2 2 mole HCl = mole H 2 (0.150 mole) L atm (273.15K) mol K = 3.36L 1.000atm

11 17. A 3.05 g sample of ammonium nitrate is introduced into an evacuated 2.18 L flask and then heated to C. What is the total pressure in the flask at C after the ammonium nitrate has completely decomposed according to the reaction below? NH 4NO 3(s) N 2O(g) + 2 H 2O(g) The moles of gas will be three times the initial moles of NH4NO3! mole NH n NH4 NO 3 = 3.05 g NH 4 NO 4 NO $ 3 3 # & = moles NH 4 NO 3 " g NH 4 NO 3 % P = nrt V =! ( moles) L atm $ # &(523.15K) " mol K % 2.18 L ( ) = 2.24 atm 18. A container with a fixed volume contains a gas sample with a pressure and temperature of atm and 25.5 C. The temperature is increased to 51.0 C. Calculate the final pressure inside the container. PV = nrt rearrange equation so properties which change are on one side of the equation. P T = nr V P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = (1.000atm)(324.15K) K = atm

12 19. A cylinder with a movable (weightless, frictionless) piston contains a sample of oxygen gas. The initial volume and temperature of the gas are L and 20.0 C respectively. The gas is heated against a fixed pressure until the new volume is L. Calculate the final temperature of the gas. PV = nrt rearrange equation so properties which change are on one side of the equation. P nr = T V T 1 V 1 = T 2 V 2 T 2 = T 1 V 2 V 1 = ( K)(2.500 L) L = K