Exam 4 Flashcards
What is the difference between heat and temperature?
Heat = total energy Temperature = average energy
When you place your hand on a light bulb that has been on for about 25 minutes, describe what happens regarding heat transfer and why it feels hot to your touch.
The light bulb is hotter than your hand so heat transfers from the light bulb to your hand since you gain heat, it feels hot.
Write the correct symbol and units for change in enthalpy.
∆H expressed in kJ
Write the correct symbol and units for change in energy.
∆E
Write the correct symbol and units for heat.
q expressed in J
Write the correct symbol and units for heat of reaction.
∆H expressed in kJ/mol
Write the correct symbol and units for change in temperature.
∆T
Write the correct symbol and units for specific heat capacity.
s = J/g°C
For the reaction 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) the heat of reaction (∆H) is -852 kJ.
Is the heat being absorbed from the surroundings or released to the surroundings?
released to the surroundings
For the reaction 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) the heat of reaction (∆H) is -852 kJ.
Is the reaction exothermic or endothermic?
exothermic
For the reaction 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) the heat of reaction (∆H) is -852 kJ.
Rewrite the equation with the heat as either a reactant or a product.
2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) + 852 kJ
For the reaction 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) the heat of reaction (∆H) is -852 kJ.
When 5.00 g of aluminum reacts, how much heat is given off?
- 00g(1 mol Al) (852 kj) = 78.9 kJ released
- ————————-
(26. 98g Al)(2 mol Al)
Is the evaporation of water an exothermic or endothermic process?
liquid to gas requires breaking bonds so process is endothermic (∆H+)
Is the formation of ice an exothermic or endothermic process?
liquid to solid requires forming bonds so process is exothermic (∆H-)
A sample of He gas occupies 375 mL at -45°C and 544 torr. What is the volume, in mL, of the gas at STP?
P₁V₁=P₂V₂→ (544 torr)(375 mL) = (760 torr)V₂→ V₂ = 321 mL
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T₁ T₂ (228.15κ) (273.15κ)
Calculate the density of SO₂ gas at 74°C and 1.46 atm.
d = P × MM → (1.46 atm)(64.065 g/mol) = 3.28 g/L
————- —————————-
RT (0.08206 1 atm/(mol×κ)(347 K)
Which gas molecules would have the highest speed at 0°C: A. carbon monoxide B. chlorine C. oxygen D. fluorine E. all have the same velocity at 0°C
A. carbon monoxide
lighter molecules move faster than heavier molecules
A sample of KIO₃ (s) is heated to produce O₂ (g) in the following reaction. The volume of oxygen gas collected is 275 mL at 45°C and has a pressure of 653 torr. How many grams of KI were formed?
2 KIO₃ (s) → 2 KI (s) +3 O₂ (g)
Find moles of O₂ first using PV = nRT
n = PV→ (653 torr)(1 atm/760 torr)(0.275 L) = 0.00905 mol O₂
—– ————————————–
RT (0.08206 L atm/(mol κ))(318 K)
Use stoichiometry to find grams of KI:
0.00905 mol O₂ (2 mol KI)(166.00 g KI) = 1.00 g KI
————————-
(3 mol O₂)(1 mol KI)
Calculate ∆H for the reaction:
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
Given the following information:
N₂ (g) + O₂ (g) → 2 NO (g) ∆H = 180.6 kJ
N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H = -91.8 kJ
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H = -483.7 kJ
2(N₂ (g) + O₂ (g)) → 2(2 NO (g)) ∆H = 2(180.6 kJ)
2(2 NH₃ (g)) → 2(N₂ (g) + 3 H₂ (g)) ∆H = 2(91.8 kJ)
3(H₂ (g) + O₂ (g)) → 3(2 H₂O (g)) ∆H = 3(-483.7 kJ)
———————————————————–
2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ∆H = 361.2 kJ
4 NH₃ (g) → 2 N₂ (g) + 6 H₂ (g) ∆H = 183.6 kJ
3 H₂ (g) + 3 O₂ (g) → 6 H₂O (g) ∆H = -1463.1 kJ
———————————————————–
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H = -906.3 kJ
Which of these substances does not have a ∆H°ƒ equal to 0.
A. O₂ (g) B. N₂ (l) C. Fe (s) D. Hg (l)
E. Br (s) F. Na (g) G. H₂ (g) H. I₂ (l)
I. Ca (s) J. Au (l) K. I (s) L. He (g)
∆H°ƒ = 0 are elemental state including diatomic
B. N₂ (l), E. Br (s), F. Na (g), H. I₂ (l), J. Au (l), K. I (s)
Calculate the ∆H for the following reaction given the ∆H°ƒ values:
IF₇ (g) + I₂ (s) → IF₅ (g) + 2 IF (g)
Substance ∆H°ƒ (kJ/mol)
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
IF (g) -95
IF₅ (g) -840
IF₇ (g) -941
∆H = (2 mol IF × -95 kJ/mol) + (1 mol IF₅ × -840 kJ/mol)− (1 mol IF₇ × -941 kJ/mol)
- ---------------------------------------------------------------------------------------
- 89 kJ/mol
An 8.55 g solid piece of aluminum is heated to 212°C and placed into a coffee cup calorimeter containing 50.0 mL of water at 25°C. The temperature change of water is recorded every 30 seconds and the highest temperature reading came to 103 °C. Calculate heat, q. Assume the density of water is 1.00 g/mL and the specific heat capacity of the water is 4.184 J/g°C.
Find q of water using q = ms∆T
(50.00g)(4.184 J/g°C)(103°C - 25°C) = 16,317.61 J
Describe the difference between potential energy and kinetic energy.
potential energy = stored energy
kinetic energy = energy of motion
Describe what we mean by conservation of energy and give an example.
Conservation of energy = energy cannot be created or destroyed, only converted from one form to another; example = in a chemical rxn, potential energy stored in chemical bonds may be converted to heat, light, or sound energy as a result of the rxn.
Explain why boiling water is an endothermic process. (HINT: Think about what is happening to the attractive forces between water molecules.)
Boiling water = endothermic because heat must be absorbed by the system in order to break the intermolecular forces that hold the molecules together.
Hydrogen gas and oxygen gas releases 482.6 kJ of heat when they combine to form steam. Is this reaction endothermic or exothermic? In which direction does heat transfer (between system and the surroundings) for this reaction? Is ∆H for this reaction positive or negative?
Exothermic; heat transfers from system to surroundings; ∆H is negative.
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H = -482.6 kJ
Interpret this thermochemical equation (i.e., how much heat is given off per amount of each substance?)
For every 2 moles of H₂, 482.6 kJ of energy are given off (482.6 kJ/2 mol H₂). This can also be written in terms of oxygen or steam: 482.6 kJ/1 mol O₂; 482.6 kJ/2 mol O₂
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H = -482.6 kJ
How much heat is released if we begin with 2.0087 g of O₂ gas?
- 0087 g O₂ (1 mol O₂)(-482.6 kJ) = -30.30 kJ
- ————————
(31. 998 g O₂)(1 mol O₂)
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H = -482.6 kJ
How much heat is released if we begin with 1.5021 g of H₂ gas?
- 5021 g H₂ (1 mol H₂)(-482.6 kJ) = -179.8 kJ
- ———————–
(2. 0158 g H₂)(2 mol H₂)
Which substance in each pair below has a higher specific heat?
A. aluminum foil or water
B. wood or metal
C. ethanol (C=2460 J/kg°C) or gold (C=129 J/kg°C)
D. mercury or copper
A. water
B. wood
C. ethanol
D. copper
How much heat is lost when a 640 g piece of copper cools from 375°C to 26°C? (The specific heat of copper is 0.385 J/kg°C)
q = sm∆T
q = (0.385J/g°C)(640g)(375°C-26°C)=85,993.6J(1kg/1000J) q = 86 kJ
The specific heat of iron is 0.4494 J/g°C. How much heat is transferred when a 24.7 g iron bar is cooled from 880°C to 13°C?
q = sm∆T
q = (0.4494 J/g°C)(24.7 kg)(1000g/1kg)(880°C-13°C) q = 9.6 x10⁶ J(1kj/1000g) q = 9.6 × 10³ J
8750 J of heat are applied to a 170 g sample of metal, causing a 56°C increase in its temperature. What is the specific heat of the metal? Which metal is it?
q = sm∆T (solve for s)
8750 = s(170g)(56°C)
8750/(170g)(56°C)
s = 0.919 J/g°C
Use the following enthalpies of reaction to determine the enthalpy for the reaction of ethylene with fluorine.
C₂H₄(g) + 6F₂(g) → 2CF₄(g) + 4HF(g) ∆H = ?
H₂(g) + F₂(g) →2HF(g) ∆H = -537 kJ
C(s) + 2F₂(g) → CF₄(g) ∆H = -680 kJ
2C(s) + 2H₂(g) → C₂H₄(g) ∆H = 52.3 kJ
2[H₂(g) + F₂(g)] → 2[2HF(g)] ∆H = 2(-537 kJ)
2[C(s) + 2F₂(g)] → 2[CF₄(g)] ∆H = 2(-680 kJ)
C₂H₄(g) → 2C(s) + 2H₂(g) ∆H = -52.3 kJ
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
2H₂(g) + 2F₂(g) → 4HF(g) ∆H = -1074 kJ
2C(s) + 4F₂(g) → 2CF₄(g) ∆H = -1360 kJ
C₂H₄(g) → 2C(s) + 2H₂(g) ∆H = -52.3 kJ
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
C₂H₄(g) + 6F₂(g) → 4HF(g) + 2CF₄(g) ∆H = -2486.3 kJ
Using the thermochemical equations below, what combination of the following numbered AH’s (1-4) will determine the ∆Hrxn? If a reaction needs to be reversed, write it as -∆H. If a reaction needs to be multiplied by a factor (x), write it as x∆H.
Mg₃N₂ + 3H₂O → 3MgO + 2NH₃ ∆Hrxn = ?
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
3Mg + N₂ → Mg₃N₂ ∆H₁
H₂ + ½O₂ → H₂O ∆H₂
Mg + ½O₂ → MgO ∆H₃
½N₂ + 3/2H₂ → NH₃ ∆H₄
Mg₃N₂ → 3Mg + N₂ -∆H₁
3H₂O → 3H₂ + 3/2O₂ -∆H₂
3Mg + 3/2O₂ → 3MgO ∆H₃
N₂ + 3H₂ → 2NH₃ ∆H₄
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
Mg₃N₂ + 3H₂O → 3MgO + 2NH₃
∆Hrxn = -∆H₁ + -3∆H₂ + 3∆H₃ + 2∆H₄
Which of the following substances do NOT have
∆H°ƒ = 0?
Cl₂ (g) Na (l) K (s) O (g) S₈ (s) Br₂ (l)
Na (l); O (g)
Calculate the standard enthalpy of formation of solid Mg(OH)₂ given the data shown below. (Hint: write the equation for the standard enthalpy of formation of Mg(OH)₂ (starting from elements and forming 1 mol of product) first.)
2Mg(s) + O₂(g) → 2MgO(s) ∆H° = -1203.6 kJ
Mg(OH)₂(s) → MgO(s) + H₂O(l) ∆H° = +37.1 kJ
2H₂(g) + O₂(g) → 2H₂O(l) ∆H° = -571.7 kJ
½[2Mg(s) + O₂(g) → 2MgO(s)] ∆H° = ½(-1203.6 kJ)
MgO(s) + H₂O(l) → 2H₂O(l) ∆H° = -(37.1 kJ)
½[H₂(g) + O₂(g) → 2H₂O(l)] ∆H° = ½(-571.7 kJ)
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
Mg(s) + ½O₂(g) → 2MgO(s) ∆H° = -601.8 kJ
MgO(s) + H₂O(l) → Mg(OH)₂(s) ∆H° = -37.1 kJ
H₂(g) + ½O₂(g) → H₂O(l) ∆H° = -285.85 kJ
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
Mg(s) + O₂(g) + H₂(g) → Mg(OH)₂(s) ∆H° = -924.75 kJ
Write the equation that represents the standard enthalpy of formation of:
a) MgO(s)
b) H₂O(l)
c) BaCl₂(s)
a) Mg(s) + ½O₂(g) → MgO(s)
b) H₂(g) + ½O₂(g) → H₂O(l)
c) Ba(s) + Cl₂(g) → BaCl₂(s)
Calculate the ∆H° of reaction for:
BaO(s) + SO₃(g) → BaSO₄(s)
The values of ∆H°ƒ are as follows: BaO(s) = -548 kj/mol; SO₃(g) = -395.7 kJ/mol; BaSO₄(s) = -1473 kJ/mol.
(∑∆H°ƒproducts) - (∑∆H°ƒreactants)
[BaSO₄] - [BaO + SO₃] =
[-1473 kJ/mol] - [-548 kJ/mol + (-395.7 kJ/mol)] =
529.3 kJ/mol
Calculate the ∆H° of reaction for:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
The values of ∆H°ƒ are as follows: C₃H₈(g) = -103.95 kJ/mol; CO₂(g) = -393.5 kJ/mol; H₂O(l) = -285.8 kJ/mol
(∑∆H°ƒproducts) - (∑∆H°ƒreactants)
[3(CO₂) + 4(H₂O)] - [C₃H₈ + 5(O₂)] =
[3(-393.5) + 4(-285.8)] - [(-103.95 + 0)] =
-2219.8 kJ
