Reviewing Thermochemistry

Chem I | Chem II AP
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1)
Write thermochemical formation reactions for the following compounds: [thermochemical = include the heat term in the reaction]
a)
AgCl(s)
+
+
kJ

Make sure to write the reaction for the formation of one mole of the given substance from its elements.

b)
CaCO3(s)
+
+
+
kJ

Make sure to write the reaction for the formation of one mole of the given substance from its elements.

2)
Write the thermochemical reaction for the combustion of one mole of C6H6(l). The combustion of 1.000 g of C6H6 produces 42 kJ of heat.
+
+
+
kJ
Calculate the ΔH for the combustion of one mole of C6H6:
1.000 g
42 kJ
=
78 g
x
 x = 3276 kJ
3)
Use bond energies to calculate:
a)
Enthalpy of formation of HF(g). Write the formation reaction first. Check to see if your answer agrees with the ΔHf° on the thermo chart.
kJ

½ H2 + ½ F2 → HF

½(435) + ½ (159) + -569 = -272 kJ
b)
ΔH for this reaction: 4 HCl(g) + O2(g) → 2 H2O(g) + 2 Cl2
kJ
4(431) + 498 + 4(-464) + 2(-243)
= 1724 + 498 - 1856 - 486
= -120 kJ
Check your answer by calculating ΔH using [products – reactants].
kJ
[2(-242) + 2(0)] - [4(-92) + 0]
= -484 - (-368)
= -116 kJ
*c)
ΔH for this reaction: I2(g) + 3 Cl2(g) → 2 ICl3(g)
kJ
151 + 3(243) + 6(-209)
= 151 + 729 - 1254
=-374 kJ

Remember that 2 ICl3 molecules have 6 I-Cl bonds

4)
H2(g) + ½ O2(g) → H2O(l)

2 Na(s) + ½ O2(g) → Na2O(s)

Na(s) + ½ O2(g) + ½ H2(g) → NaOH(s)
ΔH = - 286 kJ

ΔH = - 141 kJ

ΔH = - 425 kJ

First, rearrange the reactions by reversing the first and second reactions and doubling the third reaction. Then cancel everything that appears on both sides of the reactions. By adding the three reactions together you should get the final equation given in the problem. Do the same thing to the ΔH values given that you did to their respective reactions and add them together to get a final enthalpy change for the final reaction.

H2 + ½ O2 → H2O

2 Na + ½ O2 → Na2O

Na + ½ O2 + ½ H2 → NaOH
  H2O → H2 + ½ O2

Na2O → 2 Na + ½ O2

2 Na + O2 + H2 → 2 NaOH
  H2O → H2 + ½ O2

Na2O → 2 Na + ½ O2

2 Na + O2 + H2 → 2 NaOH
  Na2O + H2O → 2 NaOH
286 + 414 + 2(-425) = -150 kJ
Based on the information above, calculate the standard enthalpy change for the following reaction:
Na2O(s) + H2O(l) → 2 NaOH(s)
kJ
286 + 414 + 2(-425) = -150 kJ
*5)
Given:
CH4(g) + 2 O2(g) → CO2(g) + 2H2O(l)
ΔH = - 889.1 kJ
ΔHf° for H2O(l) = - 286 kJ/mole and ΔHf° for CO2(g) = - 393.5 kJ/mole
Calculate the standard heat of formation (ΔHf°) for CH4(g)
kJ
-889.1 = [-393.5 + 2(-286)] - [x + 2(0)]
-889.1 = -965.5 - x
x = ΔH = -76.4 kJ
6)
Determine the ΔH for the following reaction by using Hess’ Law of Heat Summation. Write all formation reactions. Show the addition of the reactions.
C2H4(g) + O2(g) → CO2(g) + H2O(l)
kJ

First, balance the given reaction. Then, write formation reactions for the three compounds in the given formula. Then, rearrange those reactions by reversing the C2H4 reaction and doubling the CO2 and H2O reactions. Then cancel everything that appears on both sides of the arrow and add the reactions together. Do the same thing to the ΔH's as you did to the reactions and add them together to get a final enthalpy change for the reaction.

2 C + 2 H2 → C2H4

C + O2 → CO2

H2 + ½ O2 → H2O
  C2H4 → 2 C + 2 H2

2 C + 2 O2 → 2 CO2

2 H2 + O2 → 2 H2O
  C2H42 C + 2 H2

2 C + 2 O2 → 2 CO2

2 H2 + O2 → 2 H2O
  C2H4 + 3 O2 → 2 CO2 + 2 H2O
-52 + 2(-393.5) + 2(-286) = -1411 kJ
7)
How much energy is required to raise the temperature of 1.50 kg of water from 22.0° C to steam at 100° C?
kJ

This problem will require calculating q two times. Once for heating the water and again for boiling it.

q = mcΔT = (1500 g)(4.18 J/g°C)(78°C) = 489,060 J
q = mHV = (1500 g)(2260 J/g) = 3,390,000 J
489,060 J + 3,390,000 J = 3,879,060 J x
1 kJ
1000 J
= 3880 kJ
8)
What is the specific heat of ethyl alcohol if 129 J is required to raise the temperature of 15.0 g of ethyl alcohol from 22.70° C to 26.20° C?
J/g°C
q = mcΔT 129 J = (15 g)(x)(3.5°C) x = 2.46 J/g°C
9)
How much energy is required to raise the temperature of 207 g of lead from 22.25° C to 27.65° C?
J
q = mcΔT = (207 g)(0.138 J/g°C)(5.4 °C) = 154 J
10)
If 50.0 J are added to a 32.3 g sample of nickel at 23.25° C, what is the final temperature of the nickel sample?
°C
q = mcΔT
50.0 J = (32.3 g)(0.443 J/g°C)(x - 23.25)
50.0 = 14.31x - 332.7 382.7 = 14.31x x = 26.7 °C

Alternately, you could solve for the entire ΔT term and get ΔT = 3.5°C. You could then add that to the initial temperature to get 26.75°C.

11)
Given: N2(g) + 3 H2(g) → 2 NH3(g)
ΔH = - 92.0 kJ
a)
a) How many grams of N2 are required to react with excess hydrogen to produce 250.0 kJ of energy?
g
250 kJ x
1 mol
92 kJ
x
28 g
1 mol
= 76.09 g
b)
a) How many kJ will be released if 68.0 g of NH3 are produced in this reaction?
kJ
68 g x
1 mol
17 g
x
92 kJ
2 mol
= 184 kJ
12)
How many mL of water at 25° C must be mixed with 150.0 mL of coffee at 100° C so that the resulting combination will have a temperature of 70° C? Assume that coffee and water have the same density and specific heat.
mL
mwater cwater ΔTwater = mcoffee ccoffee ΔTcoffee
(x)(4.18 J/g°C)(45°C) = (150 g)(4.18 J/g°C)(30°C) x = 100 g
Dwater = 1 g/mL so V = 100 mL
13)
A 10.0 kg piece of metal at 50.0° C is placed in 1.00 kg of water at 10.0° C. The metal and water come to the same temperature of 31.4° C. What is the specific heat of the metal?
J/g°C
mmetal cmetal ΔTmetal = mwater cwater ΔTwater
(10.0 kg)(x)(18.6°C) = (1.00 kg)(4.18 J/g°C)(21.4°C) x = 0.481 J/g°C