Heat changes accompanying Physical Processes
Understanding Heat Changes in Physical Processes
Introduction:
In the world of chemistry, understanding the heat changes accompanying physical changes, particularly in solutions, is very important. This lesson goes into the principles behind these changes, exploring concepts such as the standard heat of solution, molar heat of solution, hydration, and heat of dilution.
Heat changes accompanying the physical changes
1-Standard heat of solution ΔH0 sol
2-Standard heat of dilution ΔH0 dil
-Dissolving solute in solvent may causes:
Increasing intemperature it will be exothermic solution
Decreasing in temperature it will be eendothermic solution
How to calculate heat of solution?
q=m.cs.ΔT
m——mass = Volume in mL
Because Density of water 1g/cm3
Cs —— Specific heat of water = 4.18 J/g.oC
If volume = 1L it is called molar heat of solution
Cs —— Specific heat of water = 4.18 J/g.oC
If volume = 1L it is called molar heat of solution
Molar heat of solution:of solution:
The heat changes on dissolving one mole of solute to form one liter of solution.
Molar heat of solution= amount of heat / number of moles
ΔH= q/n
Example:
By dissolving 1mol of sulphuric acid in an amount of water to produce a
solution of 1000 ml volume, the temperature increases by 170C.
Calculate the amount of released energy
q = m .cs .ΔT
= 1000 × 4.18 × 17 = 71060 J
The source of heat of solution
What is the source of heat of solution?
1- Separating solvent molecules from each other
ΔH1 needs energy endothermic process (+ve) value
2- Separating solute molecules from each other
ΔH2 need energy endothermic process (+ve) value
3- Dissolving process (attaching solute and solvents molecules)
ΔH3 need energy endothermic process (+ve) value
ΔH0 sol=ΔH1 +ΔH2 +ΔH3
If ΔH1+ΔH2 < ΔH3 then ΔH sol. = (-ve) so its exotherrmic
If ΔH1+ΔH2 > ΔH3 then ΔH sol. = (-ve) so its endotherrmic
Hydration:
attaching of dissociated ions with water.
Standard heat of dilution ΔH0 dil :
It is the quantity of heat released or absorbed for each one mole of solute when diluting the solution from high concentration to low concentration in standard state.
Dilution process occurs in two processes:
1-Separating process ( separate solute from each other )need energy endothermic
2-Attaching process ( attaching solute to solvent ) Release energy exothermic
Quiz: Heat Changes Accompanying Physical Changes
1-What is the standard heat of solution denoted by?
a) ᅀH0 sol
b) ᅀH0 dil
c) ᅀH0 sub
d) ᅀH0 fusion
Explanation: The correct answer is (a) ᅀH0 sol. This term represents the standard heat change associated with dissolving a solute in a solvent to form a solution.
2-When dissolving a solute in a solvent, if there’s an increase in temperature, what kind of solution is formed?
a) Endothermic solution
b) Exothermic solution
c) Isobaric solution
d) Isometric solution
Explanation: The correct answer is (b) Exothermic solution. An increase in temperature indicates that heat is being released during the dissolution process, making it exothermic.
3-How is the heat of solution calculated?
a) q = m * ΔT
b) q = m * cs * ΔT
c) q = m * ΔH
d) q = n * ΔH
Explanation: The correct answer is (b) q = m * cs * ΔT.
This formula represents the heat change (q) during the dissolution process, where m is the mass of the solution, cs is the specific heat of the solvent, and ΔT is the change in temperature.
4-What does the molar heat of solution represent?
a) Heat change per mole of solute dissolved
b) Heat change per liter of solution formed
c) Heat change per gram of solute dissolved
d) Heat change per liter of solvent used
Explanation: The correct answer is (a) Heat change per mole of solute dissolved. It signifies the heat changes associated with dissolving one mole of solute to form a solution.
5-Which of the following contributes to the heat of solution?
a) Separating solvent molecules
b) Separating solute molecules
c) Dissolving process
d) All of the above
Explanation: The correct answer is (d) All of the above. The heat of solution is influenced by the energy required to separate solvent and solute molecules and the energy involved in the dissolving process itself.
6-What is hydration in the context of solutions?
a) Separating solute molecules
b) Separating solvent molecules
c) Attaching dissociated ions with water
d) Dissolving process
Explanation: The correct answer is (c) Attaching dissociated ions with water. Hydration refers to the process of solute ions being surrounded by water molecules after dissociation in a solution.
7-What does the standard heat of dilution represent?
a) Heat change during the separation of solute molecules
b) Heat change during the attachment of solute to solvent
c) Heat change during the dilution from high concentration to low concentration
d) Heat change during the dissolution process
Explanation: The correct answer is (c) Heat change during the dilution from high concentration to low concentration. It indicates the heat change associated with diluting a solution in its standard state.
8-What is the condition for a solution to be considered exothermic during dissolution?
a) ΔH1 + ΔH2 < ΔH3
b) ΔH1 + ΔH2 > ΔH3
c) ΔH1 + ΔH2 = ΔH3
d) ΔH1 + ΔH2 = 0
Explanation: The correct answer is (a) ΔH1 + ΔH2 <ΔH3. If the energy required to separate solvent and solute molecules (ΔH1 and ΔH2) is less than the energy released during the dissolving process (ᅀH3), the solution will be exothermic.
9-Which of the following is an endothermic process during dissolution?
a) Separating solvent molecules
b) Separating solute molecules
c) Dissolving process
d) None of the above
Explanation: The correct answer is (b) Separating solute molecules. When solute molecules are separated, it requires energy input, making it an endothermic process.
10-What factors affect the heat of solution?
a) Nature of solute and solvent
b) Temperature
c) Pressure
d) All of the above
Explanation: The correct answer is (d) All of the above. The heat of solution can be influenced by the nature of the solute and solvent, temperature, pressure, and other factors such as concentration and agitation.
11-How does the volume of solvent affect the molar heat of solution?
a) It has no effect
b) It decreases with increasing volume
c) It increases with increasing volume
d) It depends on the nature of the solute
Explanation: The correct answer is (c) It increases with increasing volume. Molar heat of solution is defined as the heat change when one mole of solute is dissolved to form one liter of solution, so as the volume increases, the molar heat of solution also increases.
12-Which process in dilution releases energy?
a) Separating process
b) Attaching process
c) Both a and b
d) None of the above
Explanation: The correct answer is (b) Attaching process. During dilution, the attachment of solute to solvent releases energy, making it an exothermic process.
13-What does a positive value of ΔH0 sol indicate?
a) Exothermic solution
b) Endothermic solution
c) Isobaric solution
d) Isometric solution
Explanation: The correct answer is (b) Endothermic solution. A positive value of ᅀH0 sol indicates that heat is absorbed during the dissolution process, making it endothermic.
14-How does the concentration of the solution affect the standard heat of dilution?
a) Higher concentration leads to higher heat of dilution
b) Higher concentration leads to lower heat of dilution
c) Concentration does not affect heat of dilution
d) It depends on the nature of the solute
Explanation: The correct answer is (a) Higher concentration leads to higher heat of dilution. When diluting a solution from high concentration to low concentration, more energy is released per mole of solute, resulting in a higher heat of dilution.
15-Which of the following statements is true regarding hydration?
a) It only occurs with non-ionic solutes
b) It involves the attachment of solute molecules to each other
c) It refers to the process of dissolving solute molecules in a solvent
d) It involves the attachment of dissociated ions with water molecules
Explanation: The correct answer is (d) It involves the attachment of dissociated ions with water molecules. Hydration specifically refers to the process where water molecules surround and interact with the dissociated ions of a solute in a solution.
16-What happens to the temperature of a solution if the heat of solution is negative?
a) Temperature increases
b) Temperature decreases
c) Temperature remains constant
d) It depends on the volume of the solution
Explanation: The correct answer is (a) Temperature increases. A negative heat of solution indicates an exothermic process, where heat is released into the surroundings, leading to an increase in temperature.
17-What units are typically used to express heat changes in solutions?
a) Calories
b) Joules
c) Kilograms
d) Liters
Explanation: The correct answer is (b) Joules. Heat changes in solutions are commonly expressed in joules (J) or kilojoules (kJ).
18-Which of the following factors affects the heat of solution the most?
a) Temperature
b) Nature of solute and solvent
c) Pressure
d) Volume of solvent
Explanation: The correct answer is (b) Nature of solute and solvent. The specific interactions between the solute and solvent molecules significantly influence the heat of solution.
19-What is the relationship between the heat of solution and the enthalpy change of a reaction?
a) They are equivalent terms
b) They are inversely proportional
c) They are directly proportional
d) They have no relationship
Explanation: The correct answer is (a) They are equivalent terms. The heat of solution refers to the heat change associated with dissolving a solute in a solvent, while the enthalpy change of a reaction refers to the heat change of a chemical reaction. In some cases, the heat of solution may be considered as part of the enthalpy change of a reaction.
20-Which of the following factors can affect the magnitude of the heat of dilution?
a) Temperature
b) Pressure
c) Concentration of the solution
d) All of the above
Explanation: The correct answer is (d) All of the above. The magnitude of the heat of dilution can be influenced by temperature, pressure, and the concentration of the solution being diluted.
21-How does the heat of dilution typically compare to the heat of solution?
a) Heat of dilution is always greater
b) Heat of dilution is always lesser
c) Heat of dilution can be greater or lesser depending on the solvent
d) Heat of dilution is equal to the heat of solution
Explanation: The correct answer is (b) Heat of dilution is always lesser. The heat of dilution is typically lesser in magnitude compared to the heat of solution because it involves the dilution of a solution, which generally results in less energy release or absorption compared to the process of dissolving solute in a solvent.
22-What effect does increasing the surface area of the solute have on the heat of solution?
a) Increases the heat of solution
b) Decreases the heat of solution
c) Has no effect on the heat of solution
d) Depends on the nature of the solute
Explanation: The correct answer is (a) Increases the heat of solution. Increasing the surface area of the solute allows for more interactions with the solvent molecules, leading to an increase in the heat of solution.
23-In a dilution process, what happens to the entropy of the system?
a) Increases
b) Decreases
c) Remains constant
d) Depends on the temperature
Explanation: The correct answer is (a) Increases. Dilution typically leads to an increase in disorder or randomness in the system, resulting in an increase in entropy.
24-What role does the standard state play in calculating the heat of dilution?
a) It ensures consistency in measurements
b) It defines a specific concentration and pressure for the solution
c) It standardizes the temperature to 25°C
d) It has no role in calculating the heat of dilution
Explanation: The correct answer is (b) It defines a specific concentration and pressure for the solution. The standard state provides a reference point for calculating thermodynamic properties such as the heat of dilution.
Quiz 2 :problems along with their answers related to heat changes accompanying physical changes:
Problem 1:
Calculate the heat change when dissolving 50 grams of potassium nitrate (KNO₃) in 100 grams of water if the temperature of the solution increases by 25°C. (Specific heat of water = 4.18 J/g°C)
Answer 1:
Given:
Mass of potassium nitrate (KNO₃) = 50 g
Mass of water = 100 g
Temperature change (ΔT) = 25°C
Specific heat of water (cs) = 4.18 J/g°C
Using the formula for heat change (q = m * cs * ᅀT):
q = 100 * 4.18 * 25
q = 10450 J
Therefore, the heat change when dissolving 50 grams of potassium nitrate in 100 grams of water is 10450 J.
Problem 2:
If the standard heat of solution for ammonium chloride (NH₄Cl) is -14.8 kJ/mol, calculate the heat change when dissolving 0.5 moles of NH₄Cl in 500 mL of water. Assume the molar mass of NH₄Cl is 53.49 g/mol.
Answer 2:
Given:
Standard heat of solution (ΔH0 sol) = -14.8 kJ/mol
Number of moles of NH₄Cl (n) = 0.5 moles
Volume of water = 500 mL = 0.5 L
Molar mass of NH₄Cl = 53.49 g/mol
First, calculate the amount of NH₄Cl in grams:
Mass = number of moles * molar mass
Mass = 0.5 * 53.49
Mass ≈ 26.745 g
Using the formula for heat change (ΔH = q/n):
ΔH = (-14.8 kJ/mol) * 0.5
ΔH = -7.4 kJ
Therefore, the heat change when dissolving 0.5 moles of NH₄Cl in 500 mL of water is -7.4 kJ.
Problem 3:
During the dilution of a 1 Molar solution of hydrochloric acid (HCl) from 100 mL to 500 mL, the temperature of the solution decreases by 15°C. Calculate the heat change associated with the dilution process. Assume the specific heat of water is 4.18 J/g°C.
Answer 3:
Given:
Initial volume of HCl solution = 100 mL = 0.1 L
Final volume of HCl solution = 500 mL = 0.5 L
Temperature change (ΔT) = -15°C (temperature decreases)
Specific heat of water (cs) = 4.18 J/g°C
Using the formula for heat change (q = m * cs * ΔT):
q = (0.5 – 0.1) * 1000 * 4.18 * (-15)
q ≈ -20970 J
Therefore, the heat change associated with the dilution process is approximately -20970 J.
Problem 4:
A student dissolves 0.2 moles of sodium chloride (NaCl) in 250 mL of water. If the temperature of the solution increases by 10°C, calculate the heat change. Assume the molar mass of NaCl is 58.44 g/mol and the specific heat of water is 4.18 J/g°C.
Answer 4:
Given:
Number of moles of NaCl (n) = 0.2 moles
Volume of water = 250 mL = 0.25 L
Temperature change (ΔT) = 10°C
Molar mass of NaCl = 58.44 g/mol
Specific heat of water (cs) = 4.18 J/g°C
First, calculate the mass of NaCl:
Mass = number of moles * molar mass
Mass = 0.2 * 58.44
Mass = 11.688 g
Using the formula for heat change (q = m * cs * ΔT):
q = 11.688 * 4.18 * 10
q ≈ 488.955 J
Therefore, the heat change when dissolving 0.2 moles of NaCl in 250 mL of water is approximately 488.955 J.
Problem 5:
If the standard heat of solution for calcium chloride (CaCl₂) is -82.8 kJ/mol, calculate the heat change when dissolving 2 moles of CaCl₂ in 1 liter of water. Assume the molar mass of CaCl₂ is 110.98 g/mol.
Answer 5:
Given:
Standard heat of solution (ΔH0 sol) = -82.8 kJ/mol
Number of moles of CaCl₂ (n) = 2 moles
Volume of water = 1 liter = 1 L
Molar mass of CaCl₂ = 110.98 g/mol
First, calculate the amount of CaCl₂ in grams:
Mass = number of moles * molar mass
Mass = 2 * 110.98
Mass = 221.96 g
Using the formula for heat change (ΔH = q/n):
ΔH = (-82.8 kJ/mol) * 2
ΔH = -165.6 kJ
Therefore, the heat change when dissolving 2 moles of CaCl₂ in 1 liter of water is -165.6 kJ.
Problem 6:
During the dissolution of 0.5 moles of potassium iodide (KI) in 200 mL of water, the temperature of the solution decreases by 8°C. Calculate the heat change associated with the dissolution process. Assume the specific heat of water is 4.18 J/g°C.
Answer 6:
Given:
Number of moles of KI (n) = 0.5 moles
Volume of water = 200 mL = 0.2 L
Temperature change (ΔT) = -8°C (temperature decreases)
Specific heat of water (cs) = 4.18 J/g°C
First, calculate the mass of KI:
Mass = number of moles * molar mass
Molar mass of KI = 166 g/mol (39.10 g/mol for K + 126.90 g/mol for I)
Mass = 0.5 * 166
Mass = 83 g
Using the formula for heat change (q = m * cs * ΔT):
q = 83 * 4.18 * (-8)
q ≈ -2774.64 J
Therefore, the heat change associated with the dissolution process is approximately -2774.64 J.
Problem 7:
The standard heat of dilution for a certain solute is -28 kJ/mol. Calculate the heat change when diluting 0.2 moles of the solute from 0.5 liters to 1 liter.
Answer 7:
Given:
Standard heat of dilution (ΔH0 dil) = -28 kJ/mol
Number of moles of the solute (n) = 0.2 moles
Initial volume of the solution = 0.5 liters
Final volume of the solution = 1 liter
Using the formula for heat change (ΔH = q/n):
ΔH = (-28 kJ/mol) * 0.2
ΔH = -5.6 kJ
Therefore, the heat change when diluting 0.2 moles of the solute from 0.5 liters to 1 liter is -5.6 kJ.
Problem 8:
A student dissolves 50 grams of calcium carbonate (CaCO₃) in 200 mL of water. If the temperature of the solution increases by 15°C, calculate the heat change. Assume the specific heat of water is 4.18 J/g°C.
Answer 8:
Given:
Mass of calcium carbonate (CaCO₃) = 50 grams
Volume of water = 200 mL = 0.2 L
Temperature change (ΔT) = 15°C
Specific heat of water (cs) = 4.18 J/g°C
Using the formula for heat change (q = m * cs * ΔT):
q = 50 * 4.18 * 15
q = 3135 J
Therefore, the heat change when dissolving 50 grams of calcium carbonate in 200 mL of water is 3135 J.
Problem 9:
If the standard heat of solution for sodium hydroxide (NaOH) is -44.5 kJ/mol, calculate the heat change when dissolving 1.5 moles of NaOH in 500 mL of water. Assume the molar mass of NaOH is 40.00 g/mol.
Answer 9:
Given:
Standard heat of solution (ΔH0 sol) = -44.5 kJ/mol
Number of moles of NaOH (n) = 1.5 moles
Volume of water = 500 mL = 0.5 L
Molar mass of NaOH = 40.00 g/mol
First, calculate the amount of NaOH in grams:
Mass = number of moles * molar mass
Mass = 1.5 * 40.00
Mass = 60.00 g
Using the formula for heat change (ΔH = q/n):
ΔH = (-44.5 kJ/mol) * 1.5
ΔH = -66.75 kJ
Therefore, the heat change when dissolving 1.5 moles of NaOH in 500 mL of water is -66.75 kJ.
Problem 10:
During the dilution of a 2 Molar solution of sulfuric acid (H₂SO₄) from 100 mL to 500 mL, the temperature of the solution decreases by 20°C. Calculate the heat change associated with the dilution process. Assume the specific heat of water is 4.18 J/g°C.
Answer 10:
Given:
Initial volume of H₂SO₄ solution = 100 mL = 0.1 L
Final volume of H₂SO₄ solution = 500 mL = 0.5 L
Temperature change (ΔT) = -20°C (temperature decreases)
Specific heat of water (cs) = 4.18 J/g°C
Using the formula for heat change (q = m * cs * ΔT):
q = (0.5 – 0.1) * 1000 * 4.18 * (-20)
q ≈ -16720 J
Therefore, the heat change associated with the dilution process is approximately -16720 J.
Problem 11:
Calculate the heat change when dissolving 0.3 moles of sucrose (C12H22O11) in 500 mL of water if the temperature of the solution increases by 5°C. Assume the specific heat of water is 4.18 J/g°C.
Answer 11:
Given:
Number of moles of sucrose (C12H22O11) = 0.3 moles
Volume of water = 500 mL = 0.5 L
Temperature change (ΔT) = 5°C
Specific heat of water (cs) = 4.18 J/g°C
First, calculate the mass of sucrose:
The molar mass of sucrose (C12H22O11) = 342.30 g/mol
Mass = number of moles * molar mass
Mass = 0.3 * 342.30
Mass ≈ 102.69 g
Using the formula for heat change (q = m * cs * ΔT):
q = 102.69 * 4.18 * 5
q ≈ 2149.79 J
Therefore, the heat change when dissolving 0.3 moles of sucrose in 500 mL of water is approximately 2149.79 J.
Problem 12:
If the standard heat of solution for potassium hydroxide (KOH) is -57.6 kJ/mol, calculate the heat change when dissolving 2.5 moles of KOH in 1 liter of water. Assume the molar mass of KOH is 56.11 g/mol.
Answer 12:
Given:
Standard heat of solution (ᅀH0 sol) = -57.6 kJ/mol
Number of moles of KOH (n) = 2.5 moles
Volume of water = 1 liter = 1 L
Molar mass of KOH = 56.11 g/mol
First, calculate the amount of KOH in grams:
Mass = number of moles * molar mass
Mass = 2.5 * 56.11
Mass = 140.275 g
Using the formula for heat change (ΔH = q/n):
ΔH = (-57.6 kJ/mol) * 2.5
ΔH = -144 kJ
Therefore, the heat change when dissolving 2.5 moles of KOH in 1 liter of water is -144 kJ.
Problem 13:
A student dissolves 25 grams of ammonium nitrate (NH4NO3) in 150 mL of water. If the temperature of the solution increases by 12°C, calculate the heat change. Assume the specific heat of water is 4.18 J/g°C.
Answer 13:
Given:
Mass of ammonium nitrate (NH4NO3) = 25 grams
Volume of water = 150 mL = 0.15 L
Temperature change (ΔT) = 12°C
Specific heat of water (cs) = 4.18 J/g°C
First, calculate the moles of ammonium nitrate:
The molar mass of ammonium nitrate (NH4NO3) = 80.04 g/mol
Number of moles = mass / molar mass
Number of moles = 25 / 80.04 ≈ 0.3124 moles
Using the formula for heat change (q = m * cs * ΔT):
q = 25 * 4.18 * 12
q ≈ 1254 J
Therefore, the heat change when dissolving 25 grams of ammonium nitrate in 150 mL of water is approximately 1254 J.
Problem 14:
If the standard heat of solution for magnesium sulfate (MgSO4) is -89.5 kJ/mol, calculate the heat change when dissolving 1.8 moles of MgSO4 in 500 mL of water. Assume the molar mass of MgSO4 is 120.37 g/mol.
Answer 14:
Given:
Standard heat of solution (ΔH0 sol) = -89.5 kJ/mol
Number of moles of MgSO4 (n) = 1.8 moles
Volume of water = 500 mL = 0.5 L
Molar mass of MgSO4 = 120.37 g/mol
First, calculate the amount of MgSO4 in grams:
Mass = number of moles * molar mass
Mass = 1.8 * 120.37
Mass ≈ 216.666 g
Using the formula for heat change (ΔH = q/n):
ΔH = (-89.5 kJ/mol) * 1.8
ΔH = -161.1 kJ
Therefore, the heat change when dissolving 1.8 moles of MgSO4 in 500 mL of water is -161.1 kJ.