1.
2.
(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)
3.
4.
5.
6.
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.
7.
8.
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)
9.
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.
10. 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
143
90
33
formation
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.
11. 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.
12. 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.
13. 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.
14. (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.
15. 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.
16. The combustion of carbon monoxide is represented by the equation:
CO(g)+1/2O2(g)=CO2(g)
(a) Determine the value of the standard enthalpy change for the combustion of
CO(g) at 298K using the following information:
C(s)+1/2O2(g)=CO(g) standard enthalpy change=-110.5 kJ mol−1
C(s)+ O2(g)=CO2(g) standard enthalpy change=-393.5 kJ mol−1
Reverse the first equation and add it to the second equation to obtain the
third equation.
CO(g)=C(s)+1/2O2(g) standard enthalpy change=+110.5 kJ mol−1
C(s)+ O2(g)=CO2(g) standard enthalpy change=-393.5 kJ mol−1
CO(g)+1/2O2(g)=CO2(g) standard enthalpy change=-283.0 kJ mol−1(1
point is earned for reversing the first equation, 1 point is earned for the
correct answer with sign)
OR
17.
18.
19.
20.
Enthalpy of reaction=enthalpy of formation of CO2-enthalpy of formation
of CO
Enthalpy of reaction=-393.5-(-110.5)=-283.0 kJ mol−1(2 points are earned
for determining enthalpy of reaction from the enthalpy of formation, if sign
is incorrect, only one point is earned)
Annotated solution:Know how to use Hess’s law and also know that here
because O2 has no enthalpy the enthalpy of reaction can just be calculated
from the enthalpy of formation.
2Fe(s)3/2O2(g)=Fe2O3(s) standard enthalpy of formation:-824 kJ mol−1
The reaction represented below also produces iron(III) oxide of standard enthalpy for the
reaction is 280. kJ per mole of Fe2O3(s) formed
2FeO(s)1/2O2(g)=Fe2O3(s)
(a) Calculate the standard enthalpy of formation of FeO(s)
Standard enthalpy=heat of formation of products-heat of formation of reactants
-280.=-824-2*FeO-0
FeO=-272 kJ mol−1(1 point for correct stoichiometry, 1 point for correct calculation)
Annotated solution: For this question you need to apply the formula: Standard
enthalpy=heat of formation of products-heat of formation of reactants and know that
elements have 0 enthalpy.
2 nitrogen atoms combine to form a nitrogen molecule, as represented by the following
equation:
2N(g)=N2(g)
Use the table of average bond energies below, determine the enthalpy change for the
reaction.
Bond
Average Bond Enthalpy(kJ mol−1)
N single bond N
160
N double bond N
420
N triple bond N
950
enthalpy change=-950
The reaction is exothermic because the chemical equation shows the formation of the N
triple bond N.(1 point for correct sign, 1 point for magnitude)
Annotated solution: Just make sure you know that N2 consists of one triple bond.
Consider the hydrocarbon pentane, C5H12(molar mass 72.15 g).
(a) The complete combustion of 5.00 g of pentane releases 243 kJ of heat. On
the basis of this information, calculate the value of H for the complete
combustion of one mole of pentane.
5.00g/72.15=0.0693 mol of C5H12
Enthalpy change=243/0.0693=3.51*10^3 kJ mol−1(1 point earned for
correct value of mol C5H12, 1 point earned for correct substitution and
calculation of enthalpy change (Sign required; if units given, they must be
correct))
Annotated solution: Find the moles of C5H12 and divide the enthalpy of 5g
by it.
H+(aq) + OH-(aq) → H2O(l)
A student is asked to determine the molar enthalpy of neutralization for the reaction
represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH
in an open polystyrene cup calorimeter. The heat released by the reaction is determined
by using the equation q = mc*change in T
Assume the following.
* Both solutions are at the same temperature before they are combined.
* The densities of all the solutions are the same as that of water.
* Any heat lost to the calorimeter or to the air is negligible.
* The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mc*change in T
q has units of joules (or kilojoules or calories or kilocalories)
m has units of grams or kilograms
c has units of J g-1°C-1 or J g-1K-1 (calories or kilograms acceptable
alternatives)
T has units of °C or K(1 point earned for any two units, 2 points earned for all four
units)
Annotated solution:Nothing to say. Memorize.
(b) List the measurements that must be made in order to obtain the value of q .
* volume or mass of the HCl or NaOH solutions
* initial temperature of HCl or NaOH before mixing
* final (highest) temperature of solution after mixing(1 point earned for any
volume (mass of reactant), 1 point earned for initial and final (highest) temperature
(change in T is not a measurement))
Annotated solution: Know that because you know the density so you can find the
mass or volume by finding the other. Know the formula of heat capacity
(c) Explain how to calculate The value of the molar enthalpy of neutralization for the
reaction between HCl(aq) and NaOH(aq)
Determine the quantity of the heat produced, q, from
q = mc*change in T, where m = total mass of solution; divide q by mol H2O
which would be given to determine enthalpy of neutralization:
enthalpy of neutralization=-q/mol H2O OR q/mol H2O(mol reactant can substitute
for mol H2O)(1 point earned for q, 1 point earned for enthalpy of neutralization)
Annotated solution: Just know the formula and know that here the heat is the same as
enthalpy of neutralization.
(d) The student repeats the experiment with the same equal volumes as before, but this
time uses 2.0 M HCl and 2.0 M NaOH.
(i) Indicate whether the value of q increases, decreases, or stays the same when
compared to the first experiment. Justify your prediction.
The change in T will be greater, so q increases. There are more moles
of HCl and NaOH reacting so the final temperature of the mixture
will be higher.(1 point is earned for direction and explanation)
Annotated solution: Know that temperature is dependent on kinetic energy and that
kinetic energy increases when number of molecules increase.
Note: Arguments about increased mass are not acceptable because the total mass increase
is negligible (the solutions have virtually the same density) and is not the driving force
for increases in q.
(ii) Indicate whether the value of the molar enthalpy of neutralization increases,
decreases, or stays the same when compared to the first experiment. Justify your
prediction.
Both q and mol H2O increase proportionately. However, when the quotient is
determined, there is no change in enthalpy of neutralization. Molar enthalpy is
defined as per mole of reaction, therefore it will not change when the number of
moles is doubled. (1 point is earned for direction and explanation)
Annotated solution: Self-explanatory
(e) Suppose that a significant amount of heat were lost to the air during the experiment.
What effect would this have on the calculated value of the molar enthalpy of
neutralization? Justify your answer.
Heat lost to the air will produce a smaller change in T. In the equation q = mc*change in
T a smaller change in T will produce a smaller value for q (heat released) than it should.
In the equation: enthalpy of neutralization=-q/mol H2O
the smaller magnitude of q and the constant mol H2O means that the heat of
neutralization will be less negative (more positive). (1 point is earned for direction and
explanation)
Annotated solution: Self-explanatory
Notes: change in H decreases because q decreases earns 1 point
change in T decreases because change in H decreases earns 1 point
No points earned for change in T decreases therefore q decreases