Heat changes accompanying chemical changes

Understanding Heat Changes in Chemical Reactions: A Comprehensive Guide

Introduction:

Discover the world of heat changes accompanying chemical reactions in this comprehensive guide. From understanding the principles of standard heats of formation and combustion to applying Hess’s Law for complex thermodynamic calculations, this lesson will provide you with a solid foundation in the fundamentals of thermochemistry. Dive into the world of exothermic and endothermic reactions, learn how to calculate heat changes, and explore practical applications of Hess’s Law. Let’s unravel the mysteries of heat changes in chemical transformations together!

1-Standard heat of combustion.

2- Standard heat of formation.

Combustion:
Rapid Combination between the substance and oxygen.

Heat of combustion:ΔHc

Quantity of heat released when one mole of substance completely burned in
excess amount of oxygen

Standard heat of combustion:ΔH^o _C

Quantity of heat released when one mole of substance completely burned in
excess amount of oxygen at standard conditions.

Examples:
Burning of fuel – burning of glucose inside body.

Notes:
All combustion reaction release energy so they are exothermic
(ΔH is always negative value)
– Any combustion produces CO2 & H2O

Heat of formation:ΔH_f:

Quantity of heat absorbed or released during formation of one mole of
compound from its elements.

Standard heat of formation:Δ H^o _f:

Quantity of heat released or absorbed during formation of one mole of
compound from its elements in standard conditions.

-Heat formation of element = zero
ΔH = sum of heat formation of products – Sum of heat formation of reactants

Example:
Calculate the change in the heat content of the following reaction:
CH₄ + 2 O₂ >>> CO₂ + 2H₂O
By knowing that ᅀHof of CH4 , CO2 and H2O is (-74.6 , – 393.5 , -241.8 KJ /mol) in order
ΔH =ΔH_P -ΔH_R

=[(-393.5 + (2 × – 241.8)] – [(-74.6) + (0) ]
= -802.5 KJ/mol

Relation between heat of formation and stability of the compound

Stable compounds:

-Heat content of products smaller than Heat content of reactants
-They are Exothermic compounds
-Change in Heat content has -ve value

Unstable compounds:

-Heat content of products larger than Heat content of reactant
-They are Endothermic compounds
-The change in their heat content has +ve value

Hess’s law:

Heat of reaction is constant amount in standard conditions, whether the
reaction is carried out in one step or a number of steps.

Hess’s law is used to calculate heat of reactions such as
1-very slow reactions as rust
2-Dangerous reactions
3-Some reactions that their heat changes is difficult to measure.

Example:

In the reaction:
CH₄ + 2 O₂ >>> CO₂ + 2H2O
By knowing that ᅀHof of CH4 , CO2 and H2O is (-74.6 , – 393.5 , -241.8 KJ /mol) in order
ᅀH =ᅀHP -ᅀHR

=[(-393.5 + (2 × – 241.8)] – [(-74.6) + (0) ]
= -802.5 KJ/mol

Quiz: Heat Changes Accompanying Chemical Changes

1-What is the definition of the standard heat of combustion?

a) The quantity of heat absorbed when one mole of substance completely reacts with oxygen at standard conditions.

b) The quantity of heat released when one mole of substance completely reacts with oxygen in excess at standard conditions.

c) The quantity of heat absorbed when one mole of substance is formed from its elements at standard conditions.

d) The quantity of heat released when one mole of substance is formed from its elements at standard conditions.

Explanation: The correct answer is (b). Standard heat of combustion refers to the heat released when one mole of substance completely burns in excess oxygen at standard conditions.

2-Which statement is true regarding combustion reactions?

a) All combustion reactions are endothermic.

b) Combustion reactions involve the release of energy and are exothermic.

c) Combustion reactions only produce carbon dioxide.

d) Combustion reactions do not involve the element oxygen.

Explanation: The correct answer is (b). Combustion reactions release energy and are therefore exothermic.

3-Define the standard heat of formation.

a) The quantity of heat absorbed or released during the formation of one mole of compound from its elements at standard conditions.

b) The quantity of heat released when one mole of substance completely reacts with oxygen in excess at standard conditions.

c) The quantity of heat absorbed when one mole of substance is formed from its elements at standard conditions.

d) The quantity of heat absorbed or released during the combustion of one mole of compound at standard conditions.

Explanation: The correct answer is (a). Standard heat of formation refers to the heat absorbed or released during the formation of one mole of compound from its elements at standard conditions.

4-How is the heat of a reaction calculated using heat of formation?

a) By subtracting the heat of products from the heat of reactants.

b) By adding the heat of products and the heat of reactants.

c) By dividing the heat of products by the heat of reactants.

d) By multiplying the heat of products and the heat of reactants.

Explanation: The correct answer is (a). The heat of a reaction is calculated by subtracting the heat of reactants from the heat of products.

5-Which law states that the heat of reaction is constant in standard conditions regardless of the number of steps involved in the reaction?

a) Boyle’s Law

b) Le Chatelier’s Principle

c) Hess’s Law

d) Avogadro’s Law

Explanation: The correct answer is (c). Hess’s Law states that the heat of reaction is constant in standard conditions regardless of the number of steps involved in the reaction.

6-What is the relation between the heat of formation and the stability of a compound?

a) Stable compounds have a positive heat of formation.

b) Unstable compounds have a negative heat of formation.

c) Stable compounds have a negative heat of formation.

d) Unstable compounds have a positive heat of formation.

Explanation: The correct answer is (c). Stable compounds typically have a negative heat of formation, indicating that the heat content of the products is lower than that of the reactants.

7-Which of the following statements is true regarding unstable compounds?

a) They have a positive heat of formation.

b) They are always exothermic.

c) Their heat content is lower than that of stable compounds.

d) They release energy during formation.

Explanation: The correct answer is (a). Unstable compounds generally have a positive heat of formation, meaning their heat content is higher than that of the reactants, which makes them endothermic.

8-When calculating the change in heat content using Hess’s Law, which quantities are typically involved?

a) Heat of reaction and heat of combustion

b) Heat of formation and heat of reaction

c) Heat of formation and heat of combustion

d) Heat of formation, heat of combustion, and heat of reaction

Explanation: The correct answer is (b). Hess’s Law involves the heat of formation and the heat of reaction to calculate the change in heat content.

9-In a combustion reaction, what are the typical products formed?

a) Carbon dioxide and water

b) Carbon monoxide and hydrogen

c) Oxygen and nitrogen

d) Carbon dioxide and methane

Explanation: The correct answer is (a). Combustion reactions typically produce carbon dioxide and water as products.

10-How can Hess’s Law be applied in practical situations?

a) To measure the heat of combustion of various substances

b) To determine the stability of compounds

c) To calculate the heat of formation of elements

d) To predict the products of a chemical reaction

Explanation: The correct answer is (a). Hess’s Law can be applied to measure the heat of combustion of various substances, among other practical applications.

11-Why is the standard heat of combustion preferred in thermodynamic calculations over non-standard conditions?

a) It allows for easier comparison between different substances.

b) It ensures that reactions occur at room temperature.

c) It simplifies the calculation of heat content changes.

d) It reduces the amount of oxygen required for combustion reactions.

Explanation: The correct answer is (a). The standard heat of combustion allows for easier comparison between different substances because it specifies conditions under which the combustion occurs, making the comparison more standardized.

12-Which of the following statements is true about the heat of combustion?

a) It is always positive.

b) It is the same for all substances.

c) It is the energy required to break bonds in a compound.

d) It is the energy released when a compound reacts with oxygen.

Explanation: The correct answer is (d). The heat of combustion is the energy released when a compound reacts with oxygen, typically in the form of a fire or flame.

13-How does the heat of formation of an element compare to its heat of combustion?

a) The heat of formation is always greater than the heat of combustion.

b) The heat of formation is always less than the heat of combustion.

c) The heat of formation is equal to the heat of combustion.

d) The heat of formation is unrelated to the heat of combustion.

Explanation: The correct answer is (d). The heat of formation of an element is unrelated to the heat of combustion, as the heat of formation pertains to the formation of a compound from its elements, while the heat of combustion relates to the combustion of a compound.

14-How does the standard heat of formation help determine the stability of a compound?

a) A negative heat of formation indicates a stable compound.

b) A positive heat of formation indicates a stable compound.

c) A negative heat of formation indicates an unstable compound.

d) A positive heat of formation indicates an unstable compound.

Explanation: The correct answer is (a). A negative heat of formation indicates that the compound is stable because it implies that the formation of the compound releases energy, making it thermodynamically favorable.

15-Which type of reactions is Hess’s Law particularly useful for?

a) Slow reactions involving multiple steps.

b) Reactions that occur spontaneously.

c) Reactions with highly reactive elements.

d) Reactions with low activation energies.

Explanation: The correct answer is (a). Hess’s Law is particularly useful for slow reactions involving multiple steps because it allows for the calculation of the overall heat change by considering the heats of formation of all intermediate products and reactants.

Give reasons quiz question with explanation about this lesson

Here are some quiz questions with explanations focusing on the topic of heat changes accompanying chemical changes:

Question 1: Why are combustion reactions considered exothermic?

Reasoning: Combustion reactions involve the rapid combination of a substance with oxygen, resulting in the release of energy in the form of heat and light. This release of energy indicates an exothermic process where heat is being given off to the surroundings.

Question 2: How does the heat of formation of a compound relate to its stability?

Reasoning: The heat of formation of a compound is the energy released or absorbed when one mole of the compound is formed from its elements. If the heat of formation is negative, it indicates that the formation of the compound releases energy, making it thermodynamically favorable and suggesting stability. Conversely, if the heat of formation is positive, it indicates that energy is absorbed during formation, suggesting that the compound is less stable.

Question 3: Why is Hess’s Law useful for calculating heat changes in chemical reactions?

Reasoning: Hess’s Law states that the overall heat change in a chemical reaction is constant, regardless of the pathway taken. This makes it particularly useful for calculating heat changes in reactions that may be difficult to measure directly or involve multiple steps. By using known heats of formation, Hess’s Law allows us to determine the heat change of a reaction indirectly, making it a valuable tool in thermodynamics.

Question 4: How does the standard heat of combustion differ from the standard heat of formation?

Reasoning: The standard heat of combustion refers to the heat released when one mole of a substance completely burns in excess oxygen at standard conditions. In contrast, the standard heat of formation refers to the heat released or absorbed when one mole of a compound is formed from its elements at standard conditions. While both involve heat changes accompanying chemical reactions, the standard heat of combustion specifically relates to combustion reactions, whereas the standard heat of formation pertains to the formation of compounds from their constituent elements.

Question 5: Why are stable compounds characterized by a negative heat of formation?

Reasoning: Stable compounds are those that have lower energy states and are more thermodynamically favorable. A negative heat of formation indicates that the formation of the compound releases energy, suggesting that the compound is stable. This is because the compound has a lower energy state compared to its constituent elements, making its formation favorable and indicating stability.

Question 6: How does the heat of combustion of a substance relate to its energy content?

Reasoning: The heat of combustion represents the amount of energy released when one mole of a substance undergoes complete combustion in excess oxygen. Therefore, substances with higher heats of combustion have higher energy content per mole because they release more energy when burned completely.

Question 7: Why is it important to specify standard conditions when discussing heats of formation and combustion?

Reasoning : Standard conditions ensure consistency and allow for accurate comparisons between different substances. Standard conditions typically include a temperature of 25°C and a pressure of 1 atmosphere. By specifying these conditions, scientists can compare heats of formation and combustion across different compounds under the same set of parameters.

Question 8: In a combustion reaction, why is oxygen often considered the limiting reactant?

Reasoning: In most combustion reactions, oxygen is abundant relative to the other reactants. Therefore, it is often considered the limiting reactant because the amount of oxygen present determines the extent to which the other reactants can combust. If there is insufficient oxygen available, combustion cannot proceed to completion.

Question 9: How does the heat of reaction change if a reaction is carried out in multiple steps instead of one step?

Reasoning: According to Hess’s Law, the heat of reaction is the same regardless of the pathway taken to reach the products from the reactants. Therefore, whether a reaction occurs in one step or multiple steps, the overall heat change remains constant. Hess’s Law allows scientists to calculate the overall heat change of a reaction by considering the individual steps involved.

Quiz of problems about :Heat changes accompanying chemical changes and Hess’s law

Here’s a quiz with problems and solutions focusing on heat changes accompanying chemical changes and Hess’s law:

Problem 1: Calculate the heat change for the following reaction:

2CO(g) + O2(g) → 2CO2(g)

Given the following standard heats of formation: ΔHf°(CO) = -110.5 kJ/mol, ΔHf°(CO2) = -393.5 kJ/mol, and ΔHf°(O2) = 0 kJ/mol.

Solution:

According to Hess’s Law, the overall heat change is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(CO2)] – [2ΔHf°(CO) + 1ΔHf°(O2)]
= [2(-393.5 kJ/mol)] – [2(-110.5 kJ/mol) + 0 kJ/mol]
= -787 kJ/mol – (-221 kJ/mol)
= -787 kJ/mol + 221 kJ/mol
= -566 kJ/mol

Therefore, the heat change for the reaction is -566 kJ/mol.

Problem 2: Given the following data, calculate the heat of reaction for the formation of ammonia (NH3) from its elements:

N2(g) + 3H2(g) → 2NH3(g)

ΔHf°(NH3) = -46.2 kJ/mol
ΔHf°(N2) = 0 kJ/mol
ΔHf°(H2) = 0 kJ/mol

Solution:
Using Hess’s Law, the heat of reaction can be calculated by subtracting the sum of the heats of formation of the reactants from the sum of the heats of formation of the products.

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(NH3)] – [1ΔHf°(N2) + 3ΔHf°(H2)]
= [2(-46.2 kJ/mol)] – [0 kJ/mol + 3(0 kJ/mol)]
= -92.4 kJ/mol – 0 kJ/mol
= -92.4 kJ/mol

Therefore, the heat of reaction for the formation of ammonia is -92.4 kJ/mol.

Problem 3: Given the following heats of formation, calculate the heat change for the combustion of methane (CH4) to form carbon dioxide (CO2) and water (H2O):

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

ΔHf°(CH4) = -74.6 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -241.8 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat change for the combustion reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [1ΔHf°(CO2) + 2ΔHf°(H2O)] – [1ΔHf°(CH4) + 2ΔHf°(O2)]
= [1(-393.5 kJ/mol) + 2(-241.8 kJ/mol)] – [-74.6 kJ/mol + 2(0 kJ/mol)]
= [-393.5 kJ/mol – 483.6 kJ/mol] – [-74.6 kJ/mol]
= -877.1 kJ/mol – (-74.6 kJ/mol)
= -802.5 kJ/mol

Therefore, the heat change for the combustion of methane is -802.5 kJ/mol.

Problem 4: Use Hess’s Law to find the heat of reaction for the following process:

C(s) + O2(g) → CO2(g)

Given the following data:
ΔHf°(C) = 0 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol

Solution:
Apply Hess’s Law to find the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [1ΔHf°(CO2)] – [1ΔHf°(C) + 1ΔHf°(O2)]
= [-393.5 kJ/mol] – [0 kJ/mol + 0 kJ/mol]
= -393.5 kJ/mol

Therefore, the heat of reaction for the process is -393.5 kJ/mol.

Problem 5: Calculate the heat change for the reaction:

2SO2(g) + O2(g) → 2SO3(g)

Given the following standard heats of formation:
ΔHf°(SO2) = -296.8 kJ/mol
ΔHf°(SO3) = -395.7 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat change for the reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(SO3)] – [2ΔHf°(SO2) + 1ΔHf°(O2)]
= [2(-395.7 kJ/mol)] – [2(-296.8 kJ/mol) + 0 kJ/mol]
= -791.4 kJ/mol – (-593.6 kJ/mol)
= -791.4 kJ/mol + 593.6 kJ/mol
= -197.8 kJ/mol

Therefore, the heat change for the reaction is -197.8 kJ/mol.

Problem 6: Calculate the heat of reaction for the following process using the given heats of formation:

2H2(g) + O2(g) → 2H2O(g)

Given:
ΔHf°(H2) = -286 kJ/mol
ΔHf°(O2) = 0 kJ/mol
ΔHf°(H2O) = -242 kJ/mol

Solution:
Apply Hess’s Law to find the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(H2O)] – [2ΔHf°(H2) + 1ΔHf°(O2)]
= [2(-242 kJ/mol)] – [2(-286 kJ/mol) + 0 kJ/mol]
= -484 kJ/mol – (-572 kJ/mol)
= -484 kJ/mol + 572 kJ/mol
= 88 kJ/mol

Therefore, the heat of reaction for the process is 88 kJ/mol.

Problem 7: Determine the heat of reaction for the combustion of ethane (C2H6) to form carbon dioxide (CO2) and water (H2O) using the given heats of formation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g)

Given:
ΔHf°(C2H6) = -84.7 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -241.8 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(CO2) + 3ΔHf°(H2O)] – [1ΔHf°(C2H6) + (7/2)ΔHf°(O2)]
= [2(-393.5 kJ/mol) + 3(-241.8 kJ/mol)] – [-84.7 kJ/mol + (7/2)(0 kJ/mol)]
= [-787 kJ/mol – 725.4 kJ/mol] – [-84.7 kJ/mol]
= -1512.4 kJ/mol – (-84.7 kJ/mol)
= -1512.4 kJ/mol + 84.7 kJ/mol
= -1427.7 kJ/mol

Therefore, the heat of reaction for the combustion of ethane is -1427.7 kJ/mol.

Problem 8: Use Hess’s Law to calculate the heat of reaction for the synthesis of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) gas:

N2(g) + 3H2(g) → 2NH3(g)

Given:
ΔHf°(N2) = 0 kJ/mol
ΔHf°(H2) = 0 kJ/mol
ΔHf°(NH3) = -46.2 kJ/mol

Solution:
Apply Hess’s Law to find the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(NH3)] – [1ΔHf°(N2) + 3ΔHf°(H2)]
= [2(-46.2 kJ/mol)] – [0 kJ/mol + 3(0 kJ/mol)]
= -92.4 kJ/mol – 0 kJ/mol
= -92.4 kJ/mol

Therefore, the heat of reaction for the synthesis of ammonia is -92.4 kJ/mol.

Problem 9: Determine the heat of reaction for the process using the given heats of formation:

2H2(g) + O2(g) → 2H2O(g)

Given:
ΔHf°(H2) = -286 kJ/mol
ΔHf°(O2) = 0 kJ/mol
ΔHf°(H2O) = -242 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(H2O)] – [2ΔHf°(H2) + 1ΔHf°(O2)]
= [2(-242 kJ/mol)] – [2(-286 kJ/mol) + 0 kJ/mol]
= -484 kJ/mol – (-572 kJ/mol)
= -484 kJ/mol + 572 kJ/mol
= 88 kJ/mol

Therefore, the heat of reaction for the process is 88 kJ/mol.

Problem 10: Calculate the heat of reaction for the combustion of ethane (C2H6) to form carbon dioxide (CO2) and water (H2O) using the given heats of formation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g)

Given:
ΔHf°(C2H6) = -84.7 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -241.8 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(CO2) + 3ΔHf°(H2O)] – [1ΔHf°(C2H6) + (7/2)ΔHf°(O2)]
= [2(-393.5 kJ/mol) + 3(-241.8 kJ/mol)] – [-84.7 kJ/mol + (7/2)(0 kJ/mol)]
= [-787 kJ/mol – 725.4 kJ/mol] – [-84.7 kJ/mol]
= -1512.4 kJ/mol – (-84.7 kJ/mol)
= -1512.4 kJ/mol + 84.7 kJ/mol
= -1427.7 kJ/mol

Therefore, the heat of reaction for the combustion of ethane is -1427.7 kJ/mol.

Problem 11: Determine the heat of reaction for the synthesis of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) gas:

N2(g) + 3H2(g) → 2NH3(g)

Given:
ΔHf°(N2) = 0 kJ/mol
ΔHf°(H2) = 0 kJ/mol
ΔHf°(NH3) = -46.2 kJ/mol

Solution:
Apply Hess’s Law to find the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [2ΔHf°(NH3)] – [1ΔHf°(N2) + 3ΔHf°(H2)]
= [2(-46.2 kJ/mol)] – [0 kJ/mol + 3(0 kJ/mol)]
= -92.4 kJ/mol – 0 kJ/mol
= -92.4 kJ/mol

Therefore, the heat of reaction for the synthesis of ammonia is -92.4 kJ/mol.

Problem 12: Use Hess’s Law to calculate the heat of reaction for the following process:
C(s) + O2(g) → CO2(g)

Given:
ΔHf°(C) = 0 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol

Solution:
Apply Hess’s Law to find the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [1ΔHf°(CO2)] – [1ΔHf°(C) + 1ΔHf°(O2)]
= [-393.5 kJ/mol] – [0 kJ/mol + 0 kJ/mol]
= -393.5 kJ/mol

Therefore, the heat of reaction for the process is -393.5 kJ/mol.

Problem 13: Calculate the heat of reaction for the combustion of methane (CH4) to form carbon dioxide (CO2) and water (H2O) using the given heats of formation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Given:
ΔHf°(CH4) = -74.6 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -241.8 kJ/mol

Solution:
Apply Hess’s Law to calculate the heat of reaction:

ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
= [1ΔHf°(CO2) + 2ΔHf°(H2O)] – [1ΔHf°(CH4) + 2ΔHf°(O2)]
= [1(-393.5 kJ/mol) + 2(-241.8 kJ/mol)] – [-74.6 kJ/mol + 2(0 kJ/mol)]
= [-393.5 kJ/mol – 483.6 kJ/mol] – [-74.6 kJ/mol]
= -877.1 kJ/mol – (-74.6 kJ/mol)
= -877.1 kJ/mol + 74.6 kJ/mol
= -802.5 kJ/mol

Therefore, the heat of reaction for the combustion of methane is -802.5 kJ/mol.