Good Morning! Good Morning! 04/21/2304/21/23

Today we will be… Today we will be… Preparing for tomorrow’s test Preparing for tomorrow’s test by going through the answers by going through the answers to the Practice Testto the Practice Test

Before we get into the practice Before we get into the practice test turn in your lab 19.test turn in your lab 19.

Test tomorrow.Test tomorrow.

Practice Test: Ch. 11 (Thermochemistry)Practice Test: Ch. 11 (Thermochemistry)

1.1. Complete the equation for Complete the equation for finding heat energy.finding heat energy.

H = m H = m C C TT

2.2. The specific heat capacity of The specific heat capacity of graphite is 0.71 J/(g x °C). graphite is 0.71 J/(g x °C).

Calculate the energy required Calculate the energy required to raise the temperature of 450 to raise the temperature of 450

g of graphite by 150°C.g of graphite by 150°C.

H = m H = m •• T T •• C C450g450g 0.71 0.71 JJ//g•ºCg•ºC150 150 CC

H = 47,925 J H = 47,925 J = 48 kJ = 48 kJ

3.3. It takes 770 joules of energy to It takes 770 joules of energy to raise the temperature of 50 g raise the temperature of 50 g of mercury by 110°C. What is of mercury by 110°C. What is the specific heat capacity of the specific heat capacity of

mercury?mercury?

H = m x C x Tm x Tm x T

H

m x T = C

770 J = 0.14 J/g· C

50.0 g 110C

4.4. Calculate the heat absorbed Calculate the heat absorbed by the water in a calorimeter by the water in a calorimeter

when 175 grams of copper when 175 grams of copper cools from 125.0°C to 22.0°C. cools from 125.0°C to 22.0°C. The specific heat capacity of The specific heat capacity of

copper is 0.385 J/(g x °C).copper is 0.385 J/(g x °C).

H = m H = m •• T T •• C C175g175g 0.385 0.385 JJ//g•ºCg•ºC103 103 CC

H = 6,939.62 J H = 6,939.62 J

= 6.94 kJ = 6.94 kJ

5.5. Assume 372 joules of heat are Assume 372 joules of heat are added to 5.00 g of water added to 5.00 g of water

originally at 23.0°C. What would originally at 23.0°C. What would be the final temperature of the be the final temperature of the

water? The specific heat capacity water? The specific heat capacity of water = 4.184 J/(g x °C).of water = 4.184 J/(g x °C).

H = m x C x Tm x Cm x C

H

m x C = ∆T

372 J = 17.8 C

5.00 g 4.184 J/g·C

∆Tf = 40.8 C

6.6. How much heat is required to How much heat is required to raise the temperature of 2.0 x raise the temperature of 2.0 x 102 g of aluminum by 30°C? 102 g of aluminum by 30°C? (specific heat of aluminum = (specific heat of aluminum =

0.878 J/(g x °C))0.878 J/(g x °C))

H = m H = m •• T T •• C C200g200g 0.878 0.878 JJ//g•ºCg•ºC30 30 CC

H = 5,268 J H = 5,268 J = 5.3 kJ = 5.3 kJ

7. Find the specific heat capacity of 7. Find the specific heat capacity of Lead if an 85.0 g sample of lead Lead if an 85.0 g sample of lead with an initial temperature of with an initial temperature of 99.0°C is placed into 99.5 g of 99.0°C is placed into 99.5 g of

water with an initial water with an initial temperature of 22.0 °C. The temperature of 22.0 °C. The

final temperature of the water final temperature of the water and the lead is 25.0 °C.and the lead is 25.0 °C.

First find the First find the H for water H for water H = m H = m •• T T •• C C99.599.5

gg4.184 4.184 JJ//g•ºCg•ºC 3 3 CC

H = 1,300 J H = 1,300 J

On the test I need you to show On the test I need you to show that you know how to do the that you know how to do the algebra below, but you only algebra below, but you only

have to do it oncehave to do it once

H = m x C x Tm x Tm x T

H

m x T = C

1300 J = 0.21 J/g· C

85.0 g 74C

8.8. In the following reaction: In the following reaction:

HH2 (g)2 (g) + 1/2 O + 1/2 O2 (g)2 (g) → H → H22OO(l)(l)

H = -286 kJ/mol HH = -286 kJ/mol H22

How much heat is produced How much heat is produced when 5.00g of H2 (at STP) is when 5.00g of H2 (at STP) is

reacted with excess Oreacted with excess O22? ?

5.00g 5.00g HH22 xx

1 mol 1 mol HH22

xx2.02 g 2.02 g HH22

-286 kJ-286 kJ1 mol 1 mol

HH22

= 708 = 708 kJkJ

9.9. Find the standard heat of Find the standard heat of formation for the following formation for the following

chemical reaction. Use the table chemical reaction. Use the table provided.provided.

4 NH4 NH3(g)3(g) + 5 O + 5 O2(g)2(g) 4 NO 4 NO(g)(g) + 6 + 6 HH22OO(g)(g)

H = H = HHff (products) - (products) - HHff (reactants) (reactants) HHff Prod.=(4mols Prod.=(4mols ●● 90.37 90.37 kJkJ//molmol) + (6mols ) + (6mols ●● - -

285.8 285.8 kJkJ//molmol))

HHff React. = (4mols React. = (4mols ● ● - 46.19 - 46.19 kJkJ//molmol)+(5mols )+(5mols ●● 0 0 kJkJ//molmol))

H = (361.48kJ – 1714.8kJ) – (-184.76kJ)H = (361.48kJ – 1714.8kJ) – (-184.76kJ)

H = -1168.56 kJH = -1168.56 kJ

10.10. How many joules are there in How many joules are there in 165 calories? (1 cal = 4.184 J)165 calories? (1 cal = 4.184 J)

165 cal165 calxx 4.184 4.184

JJ ==1 cal1 cal

690 690 JJ

11. 11. How much heat is absorbed How much heat is absorbed by 150.0 g of ice at - 20.0 °C by 150.0 g of ice at - 20.0 °C

to steam at 120 °C?to steam at 120 °C?Hfus= 6.01 kJ/molHfus= 6.01 kJ/mol

Hvap = 40.7 kJ/mol Hvap = 40.7 kJ/mol Cice= 2.1 J/(g °C)Cice= 2.1 J/(g °C)

Cliquid= 4.184 J/(g °C)Cliquid= 4.184 J/(g °C)Csteam= 1.7 J/(g°C)Csteam= 1.7 J/(g°C)

Hint: Calculate the energy for Hint: Calculate the energy for each phase then find the total each phase then find the total

energy.energy.

Solid phase: Solid phase: (Temperature (Temperature increase: -20.0 increase: -20.0 C to 0C to 0C)C)

H = m H = m •• T T •• C Ciceice150g150g 2.1 2.1 JJ//g•ºCg•ºC20 20 CC

H = 6,300 J H = 6,300 J = 6.3 kJ = 6.3 kJ

Melting: Melting: (Temperature stays at 0 (Temperature stays at 0 C)C)

8.32 mols 8.32 mols HH22OO xx

6.01 kJ6.01 kJ

1 mol 1 mol HH22OO

= 50 kJ= 50 kJ

150g 150g HH22OO xx

1 mol 1 mol HH22OO

==18.02 g 18.02 g HH22OO

8.32 mols 8.32 mols HH22OO

Liquid phase:Liquid phase:(Temperature increase (Temperature increase 00C to 100C to 100C)C)

H = m H = m •• T T •• C Cliquidliquid150g150g 4.184 4.184 JJ//g•ºCg•ºC100 100 CC

H = 62,800 J H = 62,800 J = 62.8 kJ = 62.8 kJ

Boiling: Boiling: (Temperature stays at 100 (Temperature stays at 100 C)C)

8.32 mols 8.32 mols HH22OO xx

40.7 kJ40.7 kJ

1 mol 1 mol HH22OO

= 333 kJ= 333 kJ

Gas phase: Gas phase: (Temperature increase 100 (Temperature increase 100 C to 120.0 C to 120.0 C)C)

H = m H = m •• T T •• C Csteamsteam150g150g 1.70 1.70 JJ//g•ºCg•ºC20 20 CC

H = 5,100 J H = 5,100 J = 5.1 kJ = 5.1 kJ

So how much energy was So how much energy was used to heat the water from used to heat the water from

-20-20CC to 120 to 120CC ? ? Warming Ice =Warming Ice = Melting Ice =Melting Ice =

Warming Water =Warming Water = Boiling Water =Boiling Water = Warming Steam Warming Steam

== Total =Total =

6.3 kJ 6.3 kJ

50.0 kJ 50.0 kJ

62.8 kJ 62.8 kJ

333.0 kJ 333.0 kJ

5.1 kJ 5.1 kJ + + 457.2 kJ 457.2 kJ

12.12. (Hess’s Law) Show All Work! 2 points(Hess’s Law) Show All Work! 2 pointsCalculate the enthalpy change Calculate the enthalpy change

((H) in kJ for the following H) in kJ for the following reaction.reaction.

2 Al2 Al(s)(s) +Fe +Fe22OO3(s) 3(s) 2 Fe 2 Fe(s)(s) +Al +Al22OO3(s)3(s)

Flip the Combustion of Fe and add them together.Flip the Combustion of Fe and add them together.

2Al2Al(s)(s) +1.5 O +1.5 O2(g)2(g) AlAl22OO3(s)3(s) H = -1669.8 kJH = -1669.8 kJ

FeFe22OO3(s)3(s) 2 Fe 2 Fe(s) (s) +1.5 O+1.5 O2(g)2(g) H = +824.2 H = +824.2 kJkJ

2 Al2 Al(s)(s) +Fe +Fe22OO3(s) 3(s) 2 Fe 2 Fe(s)(s) +Al +Al22OO3(s) 3(s)

H = -1669.8 kJ + 824.2 kJ = - 845.6 kJH = -1669.8 kJ + 824.2 kJ = - 845.6 kJ