Practice Test(Total:90 pts)
1.
A system absorbs 21.6 kJ of heat while performing 6.95 kcal of work on the surroundings.
Calculate the change in the internal energy of the system.(5 pts)
Annotated solution: Use the formula: change in internal energy = q + w. Remember to use the
same unit.
1 calory=4.184 J
Q=21.6*1000=21600 J
W=-6.95*1000*4.184=-29078.8 J(The system does the work so it’s negative)
Change in E=(21600-29078.8)/1000/4.184=-1.79 kcal
2. In which of the following processes is change in H=change in E?(3 pts)
a) Two moles of ammonia gas are cooled from 325C to 300C at 1.2 atm.
b) One gram of water is vaporized at 100C and 1 atm.
c) Two moles of hydrogen iodide gas react to form hydrogen gas and iodine gas in a 40L
container.
d) Calcium carbonate is heated to form calcium oxide and carbon dioxide in a container
with variable volume.
Annotated solution: Use the formula change in H=change in E+P*change in V, and then look
at the choices. Only C specifies the fact that its volume doesn’t change. So P*change in V is
0. So change in H=change in E. So the answer is C.
3. If, as a pioneer, you wish to warm your room by taking an object heated on top of a
pot-bellied stove to it, which of the following 15-pound objects would be the best choice?
The specific heat capacity(in J/gK) for each substance is given in parentheses. Iron(0.450),
copper(0.387), granite(0.79), gold(0.129).(10 pts)
Annotated solution: Use the formula change in H=s*m*change in T. Change in T and mass
are constant here, so if the object that has the highest specific heat capacity will yield the
highest enthalpy which is basically the warmest. So the answer is granite.
4. Calculate the enthalpy change for the reaction:
NO(g)+O(g)=NO2(g) from the following reactions:
NO(g)+O3(g)=NO2(g)+O2(g) change in H=-198.9kJ
O3(g)=1.5O2(g) change in H=-142.3kJ
O2(g)=2O(g) change in H=495.0kJ(10pts)
Annotated solution: You need to be very familiar with Hess’s Law for this question. Reverse
the second reaction and reverse the third reaction and divide it by 2.
NO(g)+O3(g)=NO2(g)+O2(g) change in H=-198.9kJ
1.5O2(g)=O3(g) change in H=142.3kJ
O(g)=1/2O2(g) change in H=-247.5kJ
Then cancel the like terms on both sides and calculate the total enthalpy change which gives
-304.1kJ.
5. Calculate the change in enthalpy for the decomposition of calcium carbonate to calcium
oxide and carbon dioxide. Given the information:
Enthalpy of formation of calcium carbonate: -1206.9kJ/mol
Enthalpy of formation of calcium oxide: -635.1kJ/mol
Enthalpy of formation of carbon dioxide: -393.5.9kJ/mol(7 pts)
Annotated solution: Write the reaction first.
CaCO3(s)=CaO(s)+CO2(g)
Use the formula: Standard enthalpy of reaction=sum of the standard enthalpies of formation
of products * moles- sum of the standard enthalpies of formation of reactants * moles. Substituate
numbers: change in enthalpy=-393.5-635.1+1206.9=178.3kJ
6. An object possesses a gravitational potential energy of 600J at a height of 10.0m. What will
its kinetic energy, potential energy and velocity be at 5.00m when it is dropped?(5 pts)
Annotated solution: Use the formulae PE=mgh, KE=1/2mv^2
600=m*10*10
M=6kg
GPE=6*10*5
GPE=300J
KE=300J(The amount PE decreases is the amount KE increases. The initial KE is 0)
300=3*v^2
V=10m/s
7. A 2.885g sample of methanol, CH3OH, is combusted in a bomb calorimeter. The temperature
of the calorimeter increases by 11.38 K. If the heat capacity of the bomb is 727.1 J/k and it
contains 1.200 kg of water. What is the heat evolved per mole of methanol combusted? The
specific heat capacity of water is 4.184 J/gK and the molar mass of methanol is 32.04
g/mol.(20 pts)
Annotated: This question is probably the hardest of the whole file. First you need to find the
heat absorbed by the calorimeter and the water by using the formula:
Q=C*change in T
Bomb calorimeter: Q=727.1*11.38=8274J
Water: Q=4.184*1.200*11.38=57137J
Total heat absorbed=8274+57137=65411J
Total heat released=-65.411kJ
Moles of methanol=2.885/32.04=0.09004mol
Heat evolved per mole of methanol combusted=-65.411/0.09004=-726.5kJ/mol
8. A sample of aluminum(0.90J/gK) absorbs 9.86J of energy while being heated from 23.2C to
30.5C. Determine the mass of the sample.(5 pts)
Annotated solution: change in T=30.5-23.2=7.3C
Use the formula: C=heat absorbed/change in T
So C=9.86/7.3=1.35J/k
1.35/0.90=1.5g
9. An iron sphere of diameter 4.00cm and density 7.86g/cm^3 at 20.0C is dropped into a
perfectly insulated beaker containing 90.0g of H2O at 50.0C(V=4/3*pi*r^3). The specific
heats of Fe and H2O are 0.45 and 4.18J/gK, respectively. What is the final temperature(in C)
of the iron sphere?(15 pts)
Annotated solution: This question looks very complicated but is actually very easy. First you
need to find the mass of the sphere by using the density and the diameter.
V=4/3*pi*2^3=33.51cm^3
D=M/V
7.86=M/33.51
M=263.39g
And then you need to be certain that the final temperatures and the enthalpy change(but
opposite in sign since one is gain and one is loss) of the iron sphere and the water are the
same since they reach thermal equilibrium. So, using the formula: change in H=s*m*change
in T, you can write
0.45*263.39*(T-20)=-4.18*90*(T-50)
T=43C
10. The heat of solution for sodium hydroxide is 44.4kJ/mol. If a 13.9g sample of NaOH
dissolves in 250.g of water at 23C, what will the final temperature of the solution be? Assume
that the solution has the same specific heat as water.(8 pts)
Annotated solution: 13.9/40=0.3475mol of NaOH
0.3475*44.4=15.429kJ of NaOH
Then use the formula change in H=s*m*change in T
15429=4.184*(250+13.9)*(Tf-23.0)
Tf=37.0C
11. For which of the species in the reaction below is the standard enthalpy of formation equal to
zero? 2Co(s)+H2(g)+8PF3=2HCo(PF3)4(l)(2 pts)
Annotated solution: Be familiar with the fact that molecules in elemental state have zero
enthalpy of formation. So the answer is 2Co and H2.
Source: The best questions from Dr Hegde’s worksheets.