Homework #6 Key
Note: Since Netscape does not have the Greek letter Delta available, I have approximated it in this document as A
4.86 A 50.0 mL volume of AgNO3 solution contains 0.0285 mol AgNO3 (silver nitrate). What is the molarity of the solution?
Answer: molarity = (moles of solute)/volume of solution) = (0.0285 mol AgNO3 )/(0.0500 L)
= 0.570 M AgNO3
4.88 A sample of oxalic acid, H2C2O4, weighing 1.274 g is placed in a 100.0 mL volumetric flask, which is then filled to the mark with water. What is the molarity of the solution?
Answer: mol H2C2O4 = (1.274 g H2C2O4)(1 mol H2C2O4/90.035 g H2C2O4)
= 0.014150 mol H2C2O4
[H2C2O4] = (0.014150 mol H2C2O4)/(.100 L) = 0.1415 M H2C2O4
4.90 How many milliliters of 0.126 M HClO4 (perchloric acid) are required to give 0.00752 mol HClO4?
Answer: volume of solution = (moles of solute)/(molarity) = (0.00752 mol)/(0.126 mol/L)
= 0.0597 L = 59.7 mL.
4.96 Describe how you would prepare 2.50 x 102 mL of 0.10 M Na2SO4. What mass (in grams) of sodium sulfate, Na2SO4, is needed?
Answer: mol Na2SO4 needed = molarity x volume = (0.10 mol Na2SO4/L) x (0.250 L)
= 0.0250 mol Na2SO4
mass Na2SO4 = (0.0250 mol Na2SO4) x (142.043 g/mol) = 3.6 g Na2SO4.
Procedure: Weigh out 3.6 g of sodium sulfate into a 250 mL volumetric flask, add enough
water to dissolve the solid, then add water to the mark on the flask and mix thoroughly.
4.106 A flask contains 53.1 mL of 0.150 M Ca(OH)2 (calcium hydroxide). How many milliliters of 0.350 M Na2CO3 (sodium carbonate) are required to react completely with the calcium hydroxide in the following reaction?
Na2CO3(aq) + Ca(OH)2(aq) --> CaCO3(s) + 2NaOH(aq)
Answer: Stoichiometry problem!
1. moles Ca(OH)2 = (0.0531 L)(0.150 mol Ca(OH)2/L) = 0.007965 mol Ca(OH)2
2. moles Na2CO3 = (0.007965 mol Ca(OH)2)(1mol Na2CO3/1 mol Ca(OH)2)
= 0.007965 mol Na2CO3
3. volume Na2CO3 solution = (0.007965 mol Na2CO3)/(0.350 mol/L) = 0.02276 L = 22.8 mL.
4.108 How many molliliters of 0.238 M KMnO4 are needed to react with 3.36 g of iron(II) sulfate, FeSO4? The reaction is as follows:
10FeSO4(aq) + 2KMnO4(aq) + 8H2SO4(aq) -->
5Fe2SO4(aq) + 2MnSO4(aq) + K2SO4(aq) +8H2O(l)
Answer: Stoichiometry problem!
1. mol FeSO4 = 3.36 g FeSO4 (1 mol FeSO4/151.909 g FeSO4) = 0.02212 mol FeSO4
2. mol KMnO4 = 0.02212 mol FeSO4 (2 mol KMnO4/10 mol KMnO4)
= 0.004424 mol KMnO4
3. volume KMnO4 solution = (0.004424 mol KMnO4)/(0.238 mol/L) = 0.0186 L = 18.6 mL.
6.8 Under what condition is the enthalpy change equalt to the heat of reaction?
Answer: When the enthalpy change occurs under constant pressure conditions.
6.10 Why is it important to give the reactants and products when giving an equation for AH?
Answer: Different physical states of the same substance have different enthaplies as ashown, for example, by the fact that solid water (ice) must be heated to form liquid water. The input of heat indicates that the enthalpy of liquid water is higher than the enthalpy of solid water. Thermochemical equations that contain substances in different physical states will, therefore, have different values for AH. By specifying the physical states of reactants and products, any ambiguity in the value of AH is removed.
6.24 Mistakenly assigned. Skip.
6.30 A bullet weighing 228 grains is moving at a speed of 2.52 x 103 ft/s. Calculate the kinetic energy of the bullet in joules and in calories. One grain equals 0.0648 g.
Answer:
K.E. = (1/2)mv2 = (1/2)(228 grains)(0.0000648 kg/grain)[(2.52 x 103 ft/s)(0.3048 m/ft)]2
= 4.36 x103 J = 4.358 x 103 J(1 cal/4.184 J) = 1.04 x 103 cal.
6.36 Nitric acid, a source of many nitrogen compounds, is produce from nitrogen dioxide. An old process for producing nitrogen dioxide employed nitrogen and oxygen.
N2(g) + 2O2(g) --> 2NO2(g)
The reaction absorbs 66.4 kJ of heat per 2 mol NO2 produced. Is the reaction endothermic or exothermic? What is the value of q?
Answer: Since the reaction absorbs heat, it must be endothermic. By absorbing heat, the energy of the chemical system is increasing and the heat must be positive in sign. Thus the value of q is +66.4 kJ.
6.42 With a platinum catalyst, ammonia will burn in oxygen to give nitric oxide, NO.
4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g) AH = -906 kJ
What is the enthalpy change for the following reaction?
NO(g) + 3/2H2O(g) --> NH3(g) + 5/4O2(g)
Answer: The latter reaction represents the former reaction reversed and multiplied through by a factor of 1/4. The AH for the latter reaction will thus have the opposite sign and be 1/4 as big as that of the former of +227 kJ.
6.48 Ethanol, C2H5OH, is mixed with gasoline and sold as gasohol. Use the following to calculate the grams of ethanol needed to provide 345 kJ of heat.
C2H5OH(l) + 3O2(g) --> 2CO2(g) + 3H2O(g) AH = -1235 kJ
Answer: A stoichiometry problem involving heat!
2. mol C2H5OH = -345 kJ(1 mol C2H5OH/-1235 kJ) = 0.2794 mol C2H5OH
3. mass C2H5OH = 0.2794 mol C2H5OH (46.069 g C2H5OH/1 mol C2H5OH)
= 12.9 g C2H5OH
6.58 Hydrazine, N2H4, is a colorless liquid used as a rocket fuel. What is the enthalpy change for the process in which hydrazine is formed from its elements?
N2(g) + 2H2(g) --> N2H4(l)
Use the following reactions and enthalpy changes:
N2H4(l) + O2(g) --> N2(g) + 2H2O(l) AH = -622.2 kJ
H2(g) + 1/2O2(g) --> H2O(l) AH = -285.8 kJ
Answer: Reverse first equation and multiply second equation by two, then add together:
N2(g) + 2H2O(l) --> N2H4(l) + O2(g) AH = +622.2 kJ
2H2(g) + O2(g) --> 2H2O(l) AH = -571.6 kJ ____________________________________________________________________________
N2(g) + 2H2(g) --> N2H4(l) AH = 50.6 kJ
6.62 Acetic acid, CH3COOH, is contained in vinegar. Suppose acetic acid was formed from its elements, according to the following equation:
2C(graphite) +2H2(g) + O2(g) --> CH3COOH(l)
Find the enthalpy change for this reaction, using the following data:
CH3COOH(l) + O2(g) --> 2CO2(g) + 2H2O(l) AH = -871 kJ
C(graphite) + O2(g) --> CO2(g) AH = -394 kJ
H2(g) + 1/2O2(g) --> H2O(l) AH = -286 kJ
Answer: Reverse first equation, multiply second equation by two, multiply third equation by 2, then add up the AH's.
2CO2(g) + 2H2O(l) --> CH3COOH(l) + O2(g) AH = +871 kJ
2C(graphite) + 2O2(g) --> 2CO2(g) AH = -788 kJ
2H2(g) + O2(g) --> 2H2O(l) AH = -572 kJ _____________________________________________________________________________
2C(graphite) +2H2(g) + O2(g) --> CH3COOH(l) AH = -489 kJ
6.68 Iron is obtained from iron ore by reduction with carbon monoxide. The overall reaction is
Fe2O3(s) +3CO(g) --> 2Fe(s) + 3CO2(g)
Calculate the standard enthalpy change for this equation. See Appendix C for data.
Answer:
AHrxn° = [3 x AHf°(CO2(g))] + [2 x AHf°(Fe(s))] - [3 x AHf°(CO(g))] - [1 x AHf°(Fe2O3(s))]
= [3 x (-393.5 kJ)] + [2 x (0 kJ)] - [3 x (-110.5 kJ)] - [1 x (-825.5 kJ)]
= -23.5 kJ
6.90 Ethylene glycol, HOCH2CH2OH, is used as antifreeze. It is produced from ethylene oxide, C2H4O, by the following reaction
C2H4O(g) + H2O(l) --> HOCH2CH2OH(l)
Use Hess's Law to obtain the enthalpy change for this reaction from the following enthalpy changes:
2C2H4O(g) + 5O2(g) --> 4CO2(g) + 4H2O(l) AH = -2612.2 kJ
HOCH2CH2OH(l) + 5/2O2(g) --> 2CO2(g) + 3H2O(l) AH = -1189.8 kJ
Answer: Divide first equation through by two, reverse second equation, then add things up.
C2H4O(g) + 5/2O2(g) --> 2CO2(g) + 2H2O(l) AH = -1306.1 kJ
2CO2(g) + 3H2O(l) --> HOCH2CH2OH(l) + 5/2O2(g) AH = +1189.8 kJ ___________________________________________________________________________________
C2H4O(g) + H2O(l) --> HOCH2CH2OH(l) AH = -116.3 kJ