Thermochemistry HOMEWORK

Lesson 1 - Homework
1. Imagine a bowl of soup in the kitchen. The soup is the system. Describe the surroundings.

The surroundings include the bowl, the air, and the entire universe around it.

2. Complete the sentence with the words endothermic/exothermic and positive/negative: When heat is absorbed by the system from the
endothermic
surroundings, the process is said to be __________________, positive
and the sign of q is ____________________.

3. Which is not true for an endothermic reaction?

a. The temperature of the surroundings decreases.
b. The heat for the reaction is positive.
c. Heat flows from the surroundings into the system.
d. The products have a lower/less energy than the reactants.
e. All of the above are true.

4. If the same amount of heat is added to 5.00 g samples of each of the metals below, which metal will experience the greatest
temperature change?
Metal Specific Heat Capacity (J/g・K) The metal with the LOWEST specific heat capacity will require the
Al 0.897 LEAST amount of energy to increase its temperature.
Sb 0.210
0.050
Cu 0.385 If the same amount of heat is added to each metal, the one with the
Fe 0.449 LOWEST specific heat capacity will reach the highest temp.
K 0.753

5. Sometimes, specific heat capacities are given in moles instead of grams, with the units J/molK. If the molar heat capacity of a compound
with the formula C2H6SO is 88.0J/molK, what is the specific heat (in J/gK) of this substance?
88.0 J/molK x (1mol/78.15g) = 1.126039667 --> 1.13 J/gK

6. A sample of aluminum (0.90J/gK) absorbs 4.32J of energy while being heated from 23.2 °C to 30.5 °C. Determine the mass of the sample.
q=mcΔT *It is unnecessary to convert the temperature to Kelvin. This is because the CHANGE in Kelvin is the same as the
4.32J = (m)(0.90)(30.5-23.2) CHANGE in Celsius. Since you're given both temperatures, you can find the CHANGE in temperature. If you were
q=0.6575342466 --> 0.66g missing one temperature, you would need to convert the single temperature into Kelvin.

7. An iron sphere with a volume of 2.3cm3 and density of 7.86g/cm3 at 20.0 °C is dropped into a perfectly insulated beaker containing 90.0g
of H2O at 100.0 °C. The specific heats of Fe and H2O are 0.450 and 4.184J/gK, respectively. What is the final temperature (in °C) of the iron
sphere? +(mcΔT) = -(mcΔT)
Heat gained = Heat lost
*Use volume and density to find grams: +[(18.078)(0.450)(Tf - (20.0+273)] = -[(90.0)(4.184)(Tf - (100.0+273)]
+q = -q
2.3cm3 x (7.86g/cm3) = 18.078g +(mcΔT) = -(mcΔT) 8.1351Tf - 2383.5843 = -376.56Tf + 140456.88
Iron = Water 384.6951Tf = 142840.4643
*We know the water will LOSE the heat because it is starting at a higher temperature. Tf = 371.3082498 ---> 371.3082498K - 273 = 98°C
8. Given that silver has a specific heat of 0.235J/g°C,
a. If a 75.0g sample of silver at 100. °C is dumped into a calorimeter (perfectly insulated container) containing 100.g of water at
20.0 °C, what will be the final temperature of the system at equilibrium? (The specific heat capacity of water is 4.184J/gK).
Heat gained = Heat lost +(mcΔT) = -(mcΔT) +(mcΔT) = -(mcΔT)
+q = -q Water = Silver +[(100.)(4.184)(Tf - (20.0+273)] = -[(75.0)(0.235)(Tf - (100.0+273)]
b. How much heat energy is transferred to the water in joules? 418.4Tf - 122591.2= -17.625Tf + 6574.125
q=mcΔT
436.025Tf = 129165.325
q=(100.)(4.184)(3=23.0-20.0)
q=1255.2 --> 1260J Tf = 296.2337595 ---> 296K - 273 = 23.0°C
9. A 55.0g sample of water at 69.0 °C is added to an unknown amount of water at 22.0 °C in a perfectly insulated container. The final
temperature is 31.0 °C. How many grams of water were there initially at 22.0 °C? (The specific heat capacity of water is 4.184J/gK).
Heat gained = Heat lost +(mcΔT) = -(mcΔT) +[(X)(4.184)(31.0 - 22.0)] = -[(55.0)(4.184)(31.0- 69.0)]
+q = -q Cold Water = Hot Water 37.656X= 8744.56
X = 232.222222g ---> 232g
10. The specific heat capacity of copper is 0.385 J/g•K. What quantity of heat is required to heat 268 g of copper from -8.2°C to +22.6°C?
(The specific heat capacity of water is 4.184J/gK).
q=mcΔT
q= (268)(0.385)(22.6-(-8.2))
q=3157.308J --> 3200J
11. A 25.5 g sample of copper at 99.8°C is dropped into a beaker containing 140 g of water at 14.5°C. What is the final temperature when
thermal equilibrium is reached? Give your answer in Celsius. (Assume no energy is lost to the surroundings, the specific heat capacity of
copper is 0.385 J/g•K, and the specific heat capacity of water is 4.184J/gK). +(mcΔT) = -(mcΔT)
+[(140)(4.184)(Tf - (14.5+273)] = -[(25.5)(0.385)(Tf - (99.8+273)]
Heat gained = Heat lost +(mcΔT) = -(mcΔT)
585.76Tf - 168406= -9.625Tf + 3659.964
+q = -q Water = Copper
595.385Tf = 172065.964
Tf = 288.9994944---> 289K - 273 = 16°C
12. A hot metal ball is dropped into a beaker containing 256 g of water, originally at 50°C. If 6.8kJ of heat is transferred from the ball to the
water, predict the final temperature of the water. (Assume no energy is lost to the surroundings. The specific heat capacity of water is
4.184J/gK).
Heat gained = Heat lost *Change 6.8kJ to J......6.8kJ x (1000J/1kJ) = 6800J
+q = -q
Water = Ball +6800J = (256)(4.184)(ΔT)
ΔT = 6.348589866 --> 6
If the original temperature was 50 and the water GAINED 6 degrees....56°C --> 60°C

13. Which gives up more heat on cooling from 50˚C to 10˚C, 50.0 g of water or 100 g of methanol? (The specific heat capacity of water is
4.184J/gK and the specific heat capacity of methanol is 2.53 J/g•K).

For Methanol: For Water:

q=(mcΔT) q=(mcΔT)
Methanol releases more heat.
q=(100)(2.53)(10-50) q=(50.0)(4.184)(10-50)
q= -10,120 --> -10,000J q= -1673.6--> -2,000J
10,000J released 2,000J released
Lesson 2- Homework
14. During a state change, temperature remains constant. For this reason, we cannot use the equation q=mcΔT to calculate energy during
state/phase changes. What equation must be used? q=mH

15. In order to use the equation above, what are the required units on the “H” value? J/g

16. Occasionally the value for "H" will be given in different units, such as kJ/mole. In that case, a conversion is necessary. Knowing that, try this
problem: A metal worker melts 65g of solid lead to make fishing sinkers. Calculate the energy change during melting. (The heat of fusion of
lead is 4.64kJ/mol). q=(0.3137065637)(4.64)
First, realize that your equation is: So, convert grams to moles:
q=1.455598456 kJ* --> 1.46kJ
q=molH 65gPb x (1mol/207.2gPb) = 0.3137065637mol
*Realize that q will be in KILOJOULES
17. A tray of water is frozen to make 278g of ice.
this time!
a. Calculate the energy change that takes place when freezing the tray of ice. (The heat of fusion of water is 334J/g).
q=mH
q=(278)(334) = 92852J --> 92,900J --> -92,900J
b. Is this energy change endothermic or exothermic? exothermic
c. Max comes home from Sam’s Club with a large amount of refrigerated cookie dough. Knowing that he won’t use the dough soon,
he decides to freeze it. Hours later, Max discovers that all of the ice cream in his freezer has melted! Why?
Freezing is an exothermic process. As the cookie dough freezes, it releases energy into the ice cream in the freezer, causing it to melt.

18. Suppose that 36 grams of ice have been added to 500.0 mL of hot tea. The original temperature of the tea is 38˚C. Heat from the tea
provides energy to melt the ice. (The heat of fusion of ice is 333J/g).

a) How much energy is used to melt the ice? Give your answer with correct sig figs and in Joules. _J.
q=mH
q=(36)(333) = 11988J --> 12,000J --> because melting is endothermic, the sign on this answer should be positive --> 12,000J

b) Assume no energy is lost to the surroundings. Find the final temperature of the tea, in Celsius, after its loss of energy to the
ice. (Use the unrounded answer from part A in your calculation). Assume
The heattheoftea has the
fusion of same 333 J/g.heat
ice isspecific capacity
Assume the as
teawater.
has the same specific
heat capacity as water). Heat Gained = Heat Lost q=mcΔT
+q = -q -11988=(500.)(4.184)(Tf - (38+273))
(ice)=(tea) -11988=2092Tf - 650612
11988=-11988 Tf = 305.2695985 --> 305.269585 - 273 = 32.26959847 --> 32°C
19. The freezing point of mercury is -38.8°C. What is the total energy change, in Joules, if 3.60g of mercury is cooled from 23.0°C to -38.8°C
and then frozen to a solid? (The specific heat capacity of mercury is 0.140 J/g•K and its heat of fusion is 11.4 J/g.)
Find the energy needed to cool the mercury down: Find the energy needed to freeze the mercury:
q=mcΔT q=mH
q=(3.60)(0.140)(-38.8-23.0) q=(3.60)(11.4)
q=-31.1472J q=41.04J --> freezing is exothermic, so change the sign --> -41.04J

Now, find the TOTAL heat: -31.1472J + -41.04J = -72.1872 --> -72.2J

20. What is the total energy change of a system that has 591J of work done on it and gives off 364J of heat? Give your answer in kJ.
Now, find the total energy change: Finally, convert to kJ:
If work is done ON a system, then w is positive: +591J q+w
Giving off heat means q is negative: -364J 227J x (1kJ/1000J) = 0.227kJ
(-364) + (+591) = +227J
21. What are the signs for q and w (work) if a system absorbs 370J of heat energy while expanding against a constant pressure? (Hint:
“Expanding against constant pressure” means that the surroundings are applying a pressure on the system (gas)….like a book/bag is
pressing on the gas doing work on it).
Work is done BY the system, so w is negative: -
Absorbing heat means q is positive: +

22. A reaction does 89kJ of work on its surroundings and absorbs 210kJ of heat in the process.
a. Is the sign on w positive or negative? negative
b. Is the sign on q positive or negative? positive
c. What is the total change in energy (ΔEnergy) for this reaction, in kJ? If negative, include the symbol - in your answer.
q+w
d. Is this reaction endothermic or exothermic? endothermic
(+210) + (-89) = +121kJ

23. Which one of the following statements is INCORRECT?

a. Energy is neither created nor destroyed in chemical reactions.
b. Kinetic energy is the energy associated with motion.
Exothermic processes RELEASE heat INTO the
c. Exothermic processes transfer heat from the surrounding into the system. surroundings, from the system.
d. Increasing the thermal energy of a gas increases the motion of its atoms.
e. Energy includes the capacity to do work.
 Lesson 3- Homework
24. What is the symbol for enthalpy, which is the energy change in a reaction that experiences a HEAT change, but does NOT do work?
ΔH
25. What are the units for q and ΔH? q is measured in J. ΔH is measured in kJ/mol.

26. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) ΔH° = -126kJ

Given the equation an enthalpy above, determine the enthalpy for the reaction shown below.
The reaction below is just the REVERSE of the reaction above. Reversing a reaction changes the SIGN on enthalpy.
+126
4NaOH(s) + O2(g) → 2Na2O2(s) + 2H2O(l) ΔH° = ________kJ

27. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) ΔH° = -126kJ

Given the equation an enthalpy above, determine the enthalpy for the reaction shown below.
The reaction below is the reaction above, but it has been REVERSED AND DIVIDED BY TWO. So change the SIGN on enthalpy and divide it by TWO.
2NaOH(s) + 1/2O2(g) → Na2O2(s) + H2O(l) ΔH° = ________kJ
+63

28. Calculate the amount of heat that is evolved (in kJ) from the reaction of 25.0g of Na 2O2 with water.
2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) ΔH° = -126kJ *It is unnecessary to include the negative sign in your answer, since
the question already stated that the heat is EVOLVED (released).
25.0g x (1mol Na2O2 / 77.98g) x (-126kJ / 2mol Na2O2) = -40.39497307kJ --> 40.4kJ
The enthalpy given with the equation is for TWO moles of Na2O2, based on the coefficient.
29. Isooctane (C8H18), one of the many hydrocarbons found in gasoline, burns according to the equation below.

2C8H18(l) + 25O2(g) → 16 CO2(g) + 18H2O(l) ΔH°rxn = -10,922 kJ

If there's 2.00 L of isooctane in your gas tank, (density = 690g/L), how much heat is evolved when it burns? (The word “evolved” means
released/given off. Also, the molar mass of isooctane is 114.26g/mol). *It is unnecessary to include the negative sign
2.00L x (690g/L) x (1mol C8H18 / 114.26g) x (-10,922kJ / 2mol C8H18) = -65956.41519 --> 66000kJ in your answer, since the question already
stated that the heat is EVOLVED (released).
30. Pick the best word to complete each blank in the following sentence:

In a calorimeter, the WATER is considered to be the _ (system, surroundings). If the temperature of the water INCREASES, then the SYSTEM
_(gained, lost) energy. This means the reaction is _(endothermic, exothermic).

31. In a coffee cup calorimeter, our equipment is made of Styrofoam, which doesn't absorb much heat. So, we assume that no heat is lost to
the cup (or heat lost to the cup is negligible) during the experiment. However, bomb calorimeters are made of metal, so they trap significant
C
heat. We must account for this heat using a calorimeter equation. Fill in the missing letters to this equation: q=________ΔT.

32. In any calorimeter calculation, we use the equation q=mcΔT. Typically, these problems provide two masses. Which mass should we use
for "m" in the equation? The mass of the water

33. If a 32.7g sample of NaOH dissolves in 150.g of water in a coffee cup calorimeter at 23.0°C, the temperature of the water will rise to
37.0°C. What is the enthalpy of solution, ΔH? (Assume the specific heat capacity of water is 4.184 J/gK and that no heat is lost to the
surroundings. The molar mass of NaOH is 40.00g/mol). Give your answer in kJ/mole, with correct sig figs, and include a negative sign if
necessary. Now, convert to kJ Then, find MOLES:
First, find the heat that went into or out of the water (surroundings):
-8786.4J x (1kJ/1000J) = -8.7864kJ 32.7g x (1mol/40.00g) = 0.8175
q=mcΔT
q=(150)(4.184)(37.0-23.0)
q= +8786.4J--> Change the SIGN to represent heat of the SYSTEM --> -8786.4J Finally, find kJ DIVIDED by moles:
Classify this reaction as endothermic or exothermic. -8.7864/0.8175 = -10.74788991--> -10.7kJ/mol
This is exothermic, because since there is a NEGATIVE sign on "q" for th system. Also, the temperature of the water (surroundings) increased.
34. If a 32.7g sample of NaCl dissolves in 150.g of water in a coffee cup calorimeter at 23.0°C, the temperature of the water will fall to
15.0°C. What is the enthalpy of solution, ΔH? (Assume the specific heat capacity of water is 4.184 J/gK and that no heat is lost to the
surroundings. The molar mass of NaCl is 58.44g/mol). Give your answer in kJ/mole, with correct sig figs, and include a negative sign if
necessary. Now, convert to kJ Then, find MOLES:
First, find the heat that went into or out of the water (surroundings): +5020.8 x (1kJ/1000J) = +5.0208kJ 32.7g x (1mol/58.44g) = 0.5595482546
q=mcΔT
q=(150)(4.184)(15.0-23.0)
q= -8786.4J --> Change the SIGN to represent heat of the SYSTEM --> +5020.8J Finally, find kJ DIVIDED by moles:
+5.0208/0.5595482546 = +8.972952661 --> +8.97kJ/mol
Classify this reaction as endothermic or exothermic.
This is endothermic, because since there is a POSITIVE sign on "q" for th system. Also, the temperature of the water (surroundings) decreased.
 *The chemicals are actually mostly water, which is the surroundings.
35. When 75.0mL of HCl is mixed with 75.0mL of NaOH (both at 23.0 °C) in a coffee cup calorimeter, the resulting solution
increases in temperature to 36.6 °C. Assuming that the solution has the density and specific heat capacity of pure water, calculate
the enthalpy of solution (in kJ/mol). of HCl. Assume each solution has a 1M concentration.
Now, convert to kJ Then, find MOLES:
First, find the heat that went into or out of the water (surroundings): -8535.36 x (1kJ/1000J) = -8.53536kJ 75.0mL x (1L/1000mL) x (1mol/1L) = 0.0750mol
q=mcΔT
q=(150)(4.184)(36.6-23.0)
q= +8535.36J --> Change the SIGN to represent heat of the SYSTEM --> -8535.36J
Finally, find kJ DIVIDED by moles:
-8.53536/0.075= -113.8048 --> -114kJ/mol

36. A 196g sample of butane, C4H10, was burned in a bomb calorimeter that had a heat capacity of 9840 J/K and contained 1500. grams of
water. The temperature of the water in the calorimeter increased from 36.84 °C to 98.82 °C. Calculate the enthalpy for the reaction shown
below. (The molar mass of butane is 58.14 g/mol). Give your answer with correct sig figs and a negative sign, if necessary.

-297 Now, find the heat that went into the metal can:
C4H10 + 6.5O2 --> 4CO2 + 5H2O ΔH = ________kJ
q=CΔT
First, find the heat that went into the water (surroundings): q=(9840)(98.82-36.84) = 609883.2J
q=mcΔT Change the SIGN to represent heat of the SYSTEM --> -609883.2J
q=(1500)(4.184)(98.82-36.84)
q= 388986.48J --> Change the SIGN to represent heat of the SYSTEM --> -388986.48J Then, find MOLES:
196g x (1mol/58.14g) = 3.371173031
Find the total heat: Now, convert to kJ
-998869.68 x (1kJ/1000J) = -998.86968kJ Finally, find kJ DIVIDED by moles:
-388986.48 + -609883.2 = -998869.68J -998.86968/3.371173031= -296.2973632--> -297kJ/mol
*Notice that there's a coefficient of ONE on butane in the equation!
37. 8.29g of sulfur (molar mass 256.48g/mol) is burned in a bomb calorimeter with excess O 2(g). The temperature increases
from 21.25°C to 26.72°C. The bomb has a heat capacity of 923 J/K and the calorimeter contains 815 g of water. Calculate the
enthalpy for the reaction shown below. Give your answer with correct sign and sig figs.
Now, find the heat that went into the metal can:
2S8(s) + 16O2(g) → 16SO2(g) ΔH = ________kJ
-1466 q=CΔT
q=(923)(26.72-21.25) = 5048.81J
First, find the heat that went into the water (surroundings): Change the SIGN to represent heat of the SYSTEM --> -5048.81JJ
q=mcΔT
q=(815)(4.184)(26.72-21.25) Then, find MOLES:
q= +18652.4812J --> Change the SIGN to represent heat of the SYSTEM --> -18652.4812J 8.29g x (1mol/256.48) = 0.0323222054
Find the total heat: Now, convert to kJ Finally, find kJ DIVIDED by moles:
-18652.4812J+ -5048.81= -23701.2912 -23701.2912 x (1kJ/1000J) = -23.7012912kJ -23.7012912/0.0323222054= -733.2819261--> -733kJ/mol

38. For each graph below, label them as endothermic or exothermic. Notice that there is a COEFFICIENT OF TWO on SULFUR in the equation!:
-733kJ x 2 = -1466kJ

Exothermic Endothermic

39. For each type of reaction below, identify the products as being HIGHER or LOWER energy than the reactants.

a. Endothermic reaction products have higher energy than the reactants

b. Exothermic reaction products have lower energy than the reactants

Lesson 4- Homework
40. Which of the equations below is not a standard molar enthalpy of formation equation?
a. CaO(s) + CO2(g) → CaCO3(s) This equation does not have ONLY pure ELEMENTS in the reactants.
b. C(s) + 1/2 O2(g) → CO(g)
c. H2(g) + 1/2 O2(g) → H2O(g)
d. N2(g) + 2 O2(g) → N2O4(g)
e. 2Na(s) + 1/2 O2(g) → Na2O(s)

41. The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. Write the equation associated with this enthalpy.
Step 1: Write the equation using ONLY pure elements in the reactants. Make sure the elements are in their standard states. Balance the equation:
N2(g) + 3H2(g) --> 2NH3(g)
Step 2: Divide ALL coefficients (if necessary) so the product has a coefficient of ONE.
1/2 N2(g) + 3/2 H2(g) --> NH3(g)
42. Using the information in the previous question, calculate the enthalpy change for producing 7.38g of NH3(g).

7.38g x (1mol/17.04g) x (-45.9kJ/1mol) = -19.87922535 --> -19.9kJ

43. Consider the following thermochemical equations:

Fe2O3 + 3CO → 2Fe + 3CO2 ΔH° = -28.0kJ Multiply this equation by 6.
3Fe +4CO2 → 4CO + Fe3O4 ΔH° = +12.5kJ Multiply this equation by 4.
Calculate the value of ΔH° (in kJ) for 6Fe2O3 + 2CO → 2CO2 + 4Fe3O4
2 2
6Fe2O3 + 18CO --> 12Fe + 18CO2 ΔH° = -168kJ
12Fe + 16CO2 --> 16CO + 4Fe3O4 ΔH° = +50.0kJ

-168 + (+50.0) = -118kJ

44. Identify all of the species in the reaction below that have an ΔH°f equal to zero.
2Co(s) + H2(g) + 8PF3 → 2HCo(PF3)4(l) Only pure elements in their standard states have a heat of formation equal to zero.

45. Calculate ΔH°rxn for 2NH3(g) + 2.5O2(g) → 2NO(g) + 3H2O(l) given

Substance ΔH°f (kJ/mol) Identify the products. Multipy their ΔH°f by their coefficients and add them together:
H2O(l) -286 (-286 x 3) + (90 x 2) = -678
NO(g) 90
NH3(g) -46 Do the same process for the reactants:
*O2(g) 0 (-46 x 2) + (0 x 2.5) = -92

Now, find PRODUCTS - REACTANTS:

*Elements in their standard states have ΔHf°equal to zero. -678 - (-92) = -586kJ

46. The information below can be found in a manual:

N2(g) + 3H2(g) →2NH3(g) ΔH° = -91.8 kJ Divide this reaction by 2.
4NH3(g) + 5O2(g) → 4 NO(g) + 6 H2O(g) ΔH° = -906.2kJ Divide this reaction by 4.
H2(g)+½ O2(g) → H2O(g) ΔH° = -241.8kJ Reverse this reaction. and multiply by 1.5
Use these values to calculate the enthalpy change for the formation of NO(g) from its elements. (This reaction is not possible to determine
in the lab).
½ N2(g) + 1/2O2(g) → NO(g) ΔH° = ?

1/2N2(g)0.5
+ 3/2H2(g) →NH3(g) ΔH° = -45.9kJ
NH3(g) + 5/4O2(g) → NO(g) + 1.5 H2O(g) ΔH° = -226.55kJ
1.5H2O(g) → 1.5H2(g) + 3/4O2(g) ΔH° = +362.7kJ

-45.9kJ + (-226.55kJ) + (+362.7kJ) = +90.25kJ

47. Write a standard formation equation, complete with its enthalpy, for the formation of P 4O10(s) from its elements (ΔHf◦[P4O10] = -2984
kJ/mol). Use this equation to calculate the energy change that would take place when 2.0g of P burns.

4P(s) + 5O2(g) --> P4O10(s) ΔHf◦ = -2984kJ

2.0g x (1mol/30.97g) x (-2984kJ/4mol) = -48.17565386--> -48kJ

48. Write a standard formation equation, complete with its enthalpy, for the formation of NO(g) from its elements (ΔH f◦[ NO] = +90.29
kJ/mol). Use this equation to calculate the energy change that would take place when 0.30 mol of NO decomposes.
1/2 N2(g) + 1/2 O2(g) --> NO(g) ΔHf◦ = +90.29 *Decomposition is the REVERSE of this reaction, so its enthalpy is NEGATIVE 90.29kJ!

0.30mol x (-90.29kJ/mol) = -27.087 --> -27kJ

49. The standard enthalpy of formation of solid barium oxide, BaO, is -553 kJ/mol, and the enthalpy of formation of barium peroxide, BaO 2,
is -634.3 kJ/mol.
a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic?
BaO(s) + ½ O2(g) → BaO2(s)
Identify the products. Multipy their ΔH°f by their coefficients and add them together:
(-634.3 x 1) = -634.3 Now, find PRODUCTS - REACTANTS:
-634.3 - (-553) = -81.3kJ
Do the same process for the reactants: *Elements in their standard states have ΔH°f equal to zero.
(-553 x 1) + (0 x 1/2) = -553
50. What does the term “standard state” mean?
It refers to the "natural" or "common" state of substance (solid,liquid, or gas) at room temperature.

51. What are the standard states of the following substances at 298K:
a) H2O liquid
b) NaCl solid *You should know these states as common facts. Also, know that MERCURY (Hg) and BROMINE (Br) are the only 2
liquid liquid elements. Finally, know the elements that are GASES: H, N, O, F, Cl and all of the noble gases.
c) Hg
d) CH4 gas

52. Given the information below, find the standard heat of formation of AlCl3.
1.5Cl2 + Fe → FeCl3 ΔH = -400kJ *Recognize this as a HEAT of FORMATION.
FeCl3 + Al → AlCl3 + Fe ΔH = +568.8kJ *Recognize this is NOT a HEAT of FORMATION.

FeCl3 + Al → AlCl3 + Fe ΔH = +568.8kJ

Heats of Formation: -400 0 X 0

*The PRODUCTS - REACTANTS of heats of formation are EQUAL to the ENTHALPY of this reaction!

Identify the products. Multipy their ΔH°f by their coefficients and add them together:
(X x 1) + (0 x 1) = X

Do the same process for the reactants:

(-400 x 1) + (0 x 1) = -400
*Elements in their standard states have ΔHf°equal to zero.
Now, set PRODUCTS - REACTANTS equal to the ENTHALPY:
(X) - (-400) = +568.8
X = +168.8kJ