- (Not sure if this is in chapter 6 or 5)Consider the three gases in the tank at 327 degrees Celsius: CH3OH(g), CO(g), and H2(g)
a)How do the average kinetic energies of the molecules of the gases compare? Explain.
The average kinetic energies are the same because all 3 gases are at the same temperature.(1 point is earned for the correct answer and explanation)
Annotated solution: The key is to be familiar with the relationship between average kinetic energy and temperature. Gases at the same temperature have same average kinetic energy.
b)Which gas has the highest molecular speed? Explain.
KE=1/2mv^2, so at a given temperature the molecules with the lowest mass have the highest average molecular speed. Therefore the molecules in H2 gas have the highest average molecular speed. (1 point is earned for the correct answer and explanation)
Annotated solution: The key is to know that lower mass means higher molecular speed and effusion rate. So hydrogen gas has the lowest mass, then it has the highest speed.
- In many organisms, glucose is oxidized to carbon dioxide and water, as represented by the following equation.
C6H12O6(s)+6O2(g)=6CO2(g)+6H2O(l)
A 2.50g sample of glucose and an excess of O2(g) were placed in a calorimeter. After the reaction was initiated and proceeded to completion, the total heat released by the reaction was calculated to be 39.0kJ.
a)Calculate the value of standard enthalpy change in kJ/mol-1, for the combustion of glucose.
2.50g*(1 mol C6H12O6/180.16g C6H12O6)=0.0139 mol C6H12O6
-39.0kJ/0.0139 mol =-2.810 kJ/mol-1(1 point is earned for the correct answer)
Annotated solution: The point here is that the enthalpy for the whole reaction is the same for reactants and products. So just find the moles of C6H12O6 first and then divide the enthalpy by the moles to find kJ/mol-1.
When oxygen is not available, glucose can be oxidized by fermentation. In that process, ethanol and carbon dioxide are produced, as represented by the following equation.
C6H12O6(s)=2C2H5OH(l)+2CO2(g) standard enthalpy change=-68.0 kJ/mol-1 at 298 K
b)Calculate the value of standard enthalpy change for the following reaction
C2H5OH(l)+3O2(g)=2CO2(g)+3H2O(g)
C6H12O6(s)+6O2(g)=6CO2(g)+6H2O(l) standard enthalpy change=-2.810 kJ/mol-1
2C2H5OH(l)+2CO2(g) = C6H12O6(s) standard enthalpy change=68.0 kJ/mol-1
2C2H5OH(l) +6O2(g) )=4CO2(g)+ 6H2O(l) standard enthalpy change=-2.740 kJ/mol-1
Thus standard enthalpy change for the reaction is -1.370 kJ/mol-1.(1 point is earned for the correct answer, 1 point is earned for the correct setup)
Annotated solution: For this question just use Hess’s law and the reactions given above and cancel out the same terms on each side and add up the total enthalpy and remember to divide by 2 at the end.
- N2(g)+3H2(g)=2NH3(g)
a)Given that standard enthalpy change for the reaction is -92.2 kJ/mol-1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain.
standard enthalpy change=bond energy of the reactants-bond energy of the products
Based on the equation above, for the standard enthalpy change to be negative, the total bond energy of the product must be larger than the total bond energy of the reactants.
OR
More energy is released as product bonds are formed than is absorbed as reactant bonds are broken. (1 point is earned for the correct answer with the correct equation and explanation)
Annotated solution: For this question just use the formula: standard enthalpy change=bond energy of the reactants-bond energy of the products and because the standard enthalpy change is negative, the products must be higher in bond energy than the reactants.
- Use principles of thermodynamics to answer the following questions
a)The gas N2O4 decomposes to form the gas NO2.
Predict the sign of standard enthalpy change for the reaction. Justify your answer.
Bonds are broken when NO2 molecules form from N2O4 molecule. Energy must be absorbed to break bonds, so the reaction is endothermic and the sign of standard enthalpy change is positive.(1 point is earned for the correct sign and the correct explanation)
Annotated solution: The only point here is that the reaction is decomposition so it needs to absorb energy so it’s endothermic and the enthalpy change is positive.
- 1/2I2(s)+1/2Cl2(g)=ICl(g) standard enthalpy change=18 kJ/mol-1
a)For the vaporization of solid iodine, I2(s)=I2(g), the value of standard enthalpy change is 62 kJ/mol-1. Using this information, calculate the standard enthalpy change for the reaction represented below.
I2(g)+Cl2(g)=2ICl(g)
I2(s)+ Cl2(g)=2ICl(g) standard enthalpy change=36 kJ/mol-1
I2(g)=I2(s) standard enthalpy change=62 kJ/mol-1
I2(g)+Cl2(g)=2ICl(g) standard enthalpy change=-26 kJ/mol-1(1 point is earned for the standard enthalpy change of either the first or second question, one point is earned for the correct sum of the standard enthalpy change values)
Annotated solution: Use Hess’s law and the reactions given and total up the enthalpy at the end and cancel the same terms on both sides.
- Answer the following questions about thermodynamics
a)Write a balanced chemical equation for the complete combustion of one mole of CH3OH(l)(standard enthalpy change=-730 kJ/mol-1). Assume products are in their standard states at 298K. Coefficients do not need to be in whole numbers.
CH3OH(l)+3/2O2(g)=CO2(g)+2H2O(l)(1 point is earned for the correct products, 1 point is earned for balancing the equation)
Annotated solution: The point here is that the combustion of hydrocarbons produce carbon dioxide and water and the rest is just balancing the equation.
b)On the basis of your answer to part (a) and the information 1/2H2(g)+O2(g)=H2O(l) standard enthalpy change=-290 kJ/mol-1 and C(s)+O2(g)=CO2(g) standard enthalpy change=-390 kJ/mol-1, determine the enthalpy change for the reaction C(s)+H2(g)+H2O(l)=CH3OH(l)
Adding the following 3 equations:
C(s)+O2(g)=CO2(g) standard enthalpy change=-390 kJ/mol-1
1/2H2(g)+O2(g)=H2O(l) standard enthalpy change=-290 kJ/mol-1
CO2(g)+2H2O(l)=CH3OH(l)+3/2O2(g) standard enthalpy change=+730 kJ/mol-1
Yields this equation: C(s)+H2(g)+H2O(l)=CH3OH(l) standard enthalpy change=+50 kJ/mol-1(1 point is earned for the correct equations, 1 point is earned for the correct value of standard enthalpy change)
Annotated solution: Use Hess’s law and the reactions given and total up the enthalpy at t he end and cancel the same terms on both sides.
c)Write the balanced chemical equation that shows the reaction that is used to determine the enthalpy of formation for one mole of CH3OH(l)
C(s)+2H2(g)+1/2O2(g)=CH3OH(l)(1 point is earned for the correct equation)
Annotated solution: The point here is that enthalpy of formation is forming a compound from its basic elements, not compounds. Then just balance.
d)On the basis of bond energies, explain why the combustion of H2(g) is exothermic.
The combustion of H2(g) is exothermic(standard enthalpy change<0) because more energy is released during the formation of 2 moles of O-H bonds than is required to break on mole of H-H bonds and one half of a mole of O-O bonds.(1 point is earned for the correct explanation)
Annotated solution: Write the reaction H2(g)+1/2O2(g)= H2O(l) first, then know that O-H is much stronger than H-H bond and O-O bond, so more energy is released than absorbed.
- Answer the following question using principles of thermodynamics.
a)If a reaction has positive standard enthalpy change, compare the sum of bond strengths of the reactants to the sum of the bond strengths of the products. Justify your answer.
Bond energy of reactants is greater than bond energy of products. Reaction is endothermic, so more energy is required to break bonds of reactants than is given off when new bonds form in products:
standard enthalpy change=bond energy of reactants-bond energy of products>0(1 point for indicating that reactants have greater bond strength, 1 point for correct explanation)
Annotated solution: Just be familiar with the formula, remember it’s reactants-products.
- In an experiment, liquid heptane, C7H16(l), is completely combusted to produce CO2(g) and H2O(l), as represented by the following equation.
C7H16(l)+11O2(g)=7CO2(g)+8H2O(l)
The heat of combustion for one mole of C7H16(l) is -4.85+*10^3kJ.
a)Using the information in the table below, calculate the value of standard enthalpy change for the formation of C7H16(l) in kJ/mol-1.
Compound / standard enthalpy change(kJ/mol-1)CO2(g) / -393.5
H2O(l) / -285.8
standard enthalpy change=bond energy of the reactants-bond energy of the products
-4850=7*(-393.5)+8*(-285.8)-C7H16-0
C7H16=-191 kJ/mol-1.(1 point for correct coefficients, 1 point for the correct substitution into the equation)
Annotated solution: Annotated solution: Just be familiar with the formula, remember it’s reactants-products. And also know that C7H16 is the product here and that compounds having only one element have no enthalpy of formation.
d) A 0.0108 mol sample of C7H16(l) is combusted in a bomb calorimeter.
(i) Calculate the amount of heat released to the calorimeter.
Heat released=0.0108*(-4850kJ)=52.4kJ of heat released(1 point for the amount of heat released)
Annotated solution: Use the molar enthalpy given and the moles given to find the heat released.
(ii) Given that the total heat capacity of its calorimeter is 9.273kJC-1, calculate the temperature of the calorimeter.
Q=Cp*change in temperature
52.4kJ=9.273kJ* change in temperature
Change in temperature=+5.65C(1 point for the correct change in temperature(must be consistent with the answer in part (d)(i)))
Annotated solution: Be familiar with the formula of heat capacity and then just plug in numbers.
- 2NO(g)+O2(g)=2NO2(g) standard enthalpy change=-114.1kJ
a)Calculate the quantity of heat released when 73.1g of NO(g) is converted to NO2(g)
73.1/30.01=2.44mol NO
2.44*114.1/2=139kJ released(1 point is earned for calculating the number of moles of NO, One point is earned for the correct answer with the correct units)
Annotated solution: Find the moles of NO first and then times it by the enthalpy and remember to divide by 2 since there are two moles NO and NO2.
b) Use the data in the table below to calculate the bond energy, in kJ/mol-1, of the N-O bond in NO2. Assume the bonds in NO2 molecule are equivalent(i.e. they have the same energy)
Bond Energy(kJ/mol-1)N-O bond in NO / 607
O-O bond in O2 / 495
N-O bond in NO2 / ?
standard enthalpy change=bond energy of the reactants-bond energy of the products
-114.1=2*607+495-4*N-O
N-O bond energy=456 kJ/mol-1(1 point can be earned if the student does the problem correctly except for indicating that only 2 N-O bonds form, 1 point can be earned if the initial equation is reversed, but the rest of the problem is done correctly)
Annotated solution: Just use the formula.
- Consider the reaction O3(g)+NO(g)=O2(g)+NO2(g)
a)Referring to the data in the table below, calculate the standard enthalpy change for the reaction at 25C, be sure to show your work.
O3(g) / NO(g) / NO2(g)standard enthalpy of formation / 143 / 90 / 33
standard enthalpy change=33-(90+143)=-200kJ(correct setup(work), a numerical result, and a negative sign, earn this point, no math errors point deducted for computational mistakes)
Annotated solution: Just use the formula.
- Hydrogen gas burns in air according to the reaction below
2 H2(g) + O2(g) → 2 H2O(l)
(a) Calculate the standard enthalpy change, for the reaction represented by the equation above. (The molar enthalpy of formation for H2O(l) is −285.8 kJ mol−1 at 298 K.)
molar enthalpy of formation=[2 (−285.8)] − [2(0) + 1(0)] = −571.6 kJ mol−1(1 point is earned for the correct answer)
Annotated solution: Just remember the formula:molar enthalpy of formation=enthalpy of the products-enthalpy of the reaction and that elements have 0 enthalpy.
(b) Calculate the amount of heat, in kJ, that is released when 10.0 g of H2(g) is burned in air.
Q=10/2.016*285.8=1.42*10^3kJ(1 point is earned for the correct setup, 1 point is earned for the correct answer)
Annotated solution: Just find the moles first and times it by the molar enthalpy.
(c) Given that the molar enthalpy of vaporization for H2O(l) is 44.0 kJ mol−1 at 298 K, what is the standard enthalpy change for the reaction 2 H2(g) + O2(g) → 2 H2O(g) ?
2 H2(g) + O2(g) → 2 H2O(l) −571.6 kJ
2 H2O(l) → 2 H2O(g) +2(44.0) kJ
2 H2(g) + O2(g) → 2 H2O(g) −483.6 kJ(1 point is earned for the correct answer)
Annotated solution: Just use Hess’s Law and total the enthalpy and cancel out the same terms to get the final equation.
- Indicate whether the statement: The large negative ΔH° for the combustion of hydrazine results from the large release of energy that occurs when the strong bonds of the reactants are broken is true or false. Justify your answer.
The statement is false on two counts. First, energy is released not when bonds are broken, but rather when they are formed. Second, the bonds in the reactants are relatively weak compared to the bonds in the products.(1 point is earned for correctly identifying the statement as false along with a valid justification)
Annotated solution: Be familiar with the fact that bonds broken absorbs energy and bonds form release energy and bonds are only broken when the bonds in the products are more stable and thus are thus stronger.
- A student performs an experiment to determine the molar enthalpy of solution of urea, H2NCONH2. The student places 91.95 g of water at 25°C into a coffee cup calorimeter and immerses a thermometer in the water. After 50 s, the student adds 5.13 g of solid urea, also at 25°C, to the water and measures the temperature of the solution as the urea dissolves. A plot of the temperature data is shown in the graph below.
(a) Determine the change in temperature of the solution that results from the dissolution of the urea.
Change in temperature=21.8-25.0=-3.2 degrees Celsius(One point is earned for the correct temperature change)
Annotated solution: Just look at the graph.
(b) According to the data, is the dissolution of urea in water an endothermic process or an exothermic process? Justify your answer.
The process is endothermic. The decrease in temperature indicates that the process for the dissolution of urea in water requires energy.( One point is earned for the correct choice with justification)
Annotated solution: Temperature decreases so it means heat is absorbed by the system so it’s endothermic.
(c) Assume that the specific heat capacity of the calorimeter is negligible and that the specific heat capacity of the solution of urea and water is 4.2 J g−1 °C−1 throughout the experiment.
(i) Calculate the heat of dissolution of the urea in joules.
Assuming that no heat energy is lost from the calorimeter and given that the calorimeter has a negligible heat capacity, the sum of the heat of dissolution, qsoln and the change in heat energy of the urea-water mixture must equal zero.
qsoln + mcΔT = 0 ⇒qsoln = − mcΔT
msoln = 5.13 g + 91.95 g = 97.08 g
qsoln = −(97.08 g)(4.2 J g −1°C−1)(−3.2°C) = 1.3 × 10^3 J(One point is earned for the correct setup. One point is earned for the correct numerical result for the heat of dissolution)
Annotated solution: Remember the formula: q=mcΔT and because calorimeter has no heat capacity, just plug in numbers.
(ii) Calculate the molar enthalpy of solution of urea in kJ mol−1.
molar enthalpy of solution=heat/moles of solute
molar mass of urea=4*1+2*14+12+16=60.0g mol−1
moles of urea=5.13 g urea/60g=0.0855mol
molar enthalpy of solution=1.3*10^3/0.0855mol=15 kJ mol−1(One point is earned for the calculation of moles of urea. One point is earned for the correct numerical result with correct algebraic sign. )
Annotated solution: Find the moles and divide the heat energy by it.
(f) The student performs a third trial of the experiment but this time adds urea that has been taken directly from a refrigerator at 5°C. What effect, if any, would using the cold urea instead of urea at 25°C have on the experimentally obtained value of molar enthalpy of solution? Justify your answer.
There would be an increase in the obtained value for molar enthalpy of solution because the colder urea would have caused a larger negative temperature change.( One point is earned for the correct prediction with justification)
Annotated solution: Common sense.
- (a) For reaction Y at 298 K, which is larger: the total bond energy of the reactants or the total bond energy of the products if the enthalpy change is positive? Explain.
The total bond energy of the reactants is larger. Reaction Y is endothermic, so there is a net input of energy as the reaction occurs. Thus, the total energy required to break the bonds in the reactants must be greater than the total energy released when the bonds are formed in the products.( One point is earned for the correct answer with appropriate explanation)
Annotated solution: Similar question as before, same idea
(b) Is the following statement true or false? Justify your answer.
“On the basis of the data in the table, it can be predicted that reaction Y will occur more rapidly than reaction X will occur.”
The statement is false.
Thermodynamic data for an overall reaction have no bearing on how slowly or rapidly the reaction occurs.( One point is earned for the correct answer with appropriate justification)
Annotated solution: This is just a fact. Remember it.
- The following questions relate to the synthetic reaction:
N2(g)+3F2(g)=2NF3(g) standard enthalpy change=-264 kJ mol−1
(a)Calculate the standard enthalpy change that occurs when a 0.256 mol sample of NF3(g) is formed from N2(g) and F2(g) at 1.00 atm and 298K.
0.256*(-264)/2=33.8KJ(1 point is earned by multiply the standard enthalpy change by the number of moles of NF3(g) formed, 1 point is earned by recognizing that 2.00 mol of NF3(g) are produced by the reaction as it is written, 1 point is earned for the correct answer including kJ or J)
Annotated solution: Basically, just be careful that 2 moles of product formed so in order to find the molar enthalpy change you have divide the given data by 2.
The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and bond formation in the product.
(b)How many bonds are formed when two molecules of NF3 are produced according to the equation in the box above?
There are six N-F bonds formed.(1 point is earned for the correct answer)
Annotated solution: Just know that NF3 has three single bonds
(c)Use both the information above and the table of average bond enthalpies below to calculate the average enthalpy of the F-F bond.
Bond / Average Bond Enthalpy(kJ mol−1)N triple bond N / 946
N-F / 272
F-F / ?
Enthalpy change=enthalpy of bonds broken-enthalpy of bonds formed
-264=946+3*F-F-6*272
F-F=141 kJ mol−1(1 point is earned for the correct number of bonds in all three compounds multiplied by the average bond enthalpies, 1 point is earned for the correct answer including kJ or J;Note: A total of one point is earned if an incorrect number of bonds is substituted in a correct equation and the answer is reasonable(i.e.positive))
Annotated solution: Just be sure that you recognize that F2 consists of one single bond and N2 has one triple bond. Then just plug the numbers into the formula multiplied by the number of bonds.
- The combustion of carbon monoxide is represented by the equation:
CO(g)+1/2O2(g)=CO2(g)