Thermodynamics is the scientific study of the transformation of heat into other forms of energy, such as work.
Key Concepts:
A system is a portion of the universe that is isolated for study or analysis.
Surroundings: Everything outside the system.
Internal Energy (U): The total energy of a system, including kinetic and potential energy.
Enthalpy measures the system’s heat content under constant pressure.
The enthalpy change of a system is equal to the heat absorbed or released at constant pressure.
Entropy (S)
Second Law of Thermodynamics: The total entropy of the universe always increases in any spontaneous process.
Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero (0 K) is zero.
Thermodynamic Processes:
Isothermal: Temperature remains constant.
Adiabatic: No heat exchange with the surroundings.
Isobaric: Pressure remains constant.
Isochoric: Volume remains constant.
Thermodynamic Properties:
Enthalpy of Formation: The enthalpy change when one mole of a compound is formed from its elements in their standard states.
Bond enthalpy is the energy needed to break a chemical bond.
Standard Enthalpy of Combustion: The enthalpy change when one mole of a substance is burned in oxygen under standard conditions.
Applications of Thermodynamics:
Chemical reactions: Understanding the energy changes involved in chemical reactions.
Engineering: Designing efficient engines and machines.
Environmental science: Studying energy flow in ecosystems.
Thermodynamics provides a framework for understanding the behavior of matter and energy, and its applications are widespread in various fields.
1. Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Ans : The correct answer is:
(ii) whose value is independent of path
A thermodynamic state function is a property of a system that depends only on its current state and not on the path taken to reach that state.
This means that the value of a state function does not depend on the history of the system, only on its current conditions.
2 .For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0
Ans : The correct answer is (iii) q = 0.
An adiabatic process is one with no heat exchange.. This means that the value of q, which represents heat, is zero.
“The other options are incorrect because…”
(i) ∆T = 0: This condition represents an isothermal process, where the temperature remains constant.
(ii) ∆p = 0: This condition represents an isobaric process, where the pressure remains constant.
(iv) w = 0: This condition represents a process where no work is done on or by the system.
Therefore, the condition for an adiabatic process is q = 0.
3. The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Ans : The correct answer is (ii) zero
The standard state of an element is its most stable form at a specified temperature and pressure (usually 25°C and 1 atm). By definition, the enthalpy of an element in its standard state is zero. This means that no energy is required to form the element from its constituent atoms under standard conditions.
4. ∆U of combustion of methane is – X kJ mol–1. The value of ∆H is
(i) = ∆U
(ii) > ∆U
(iii) < ∆U
(iv) = 0
Ans : The enthalpy change (∆H) and the internal energy change (∆U) for a reaction are related by the equation:
∆H = ∆U + nRT
where:
n is the number of moles of gas produced in the reaction
R is the gas constant
T is the temperature in Kelvin
“The balanced equation for methane combustion is:”
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
In this reaction, the number of moles of gas decreases from 3 (1 mole of CH₄ + 2 moles of O₂) to 1 (1 mole of CO₂). Therefore, n = -2.
Since the reaction is exothermic (releases heat), ∆H is negative. This means that ∆U must also be negative, and the magnitude of ∆H will be less than the magnitude of ∆U due to the negative value of nRT.
Therefore, the correct answer is (iii) < ∆U.
The enthalpy change (∆H) for the combustion of methane is less than the internal energy change (∆U) due to the decrease in the number of moles of gas during the reaction.
5 .The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be
(i) –74.8 kJ mol–1
(ii) –52.27 kJ mol–1
(iii) +74.8 kJ mol–1
(iv) +52.26 kJ mol–1.
Ans : The enthalpy of formation (ΔHf) of a compound is the enthalpy change when one mole of the compound is formed from its elements in their standard states.
For methane (CH₄), the formation reaction can be written as:
C(graphite) + 2H₂(g) → CH₄(g)
The enthalpy of formation of CH₄ can be calculated using Hess’s law, which states that the enthalpy change of a reaction is independent of the pathway taken. We can use the enthalpies of combustion of methane, graphite, and dihydrogen to calculate the enthalpy of formation of CH₄.
The balanced chemical equations for the combustion reactions are:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890.3 kJ/mol
C(graphite) + O₂(g) → CO₂(g) ΔH = -393.5 kJ/mol
H₂(g) + 1/2O₂(g) → H₂O(l) ΔH = -285.8 kJ/mol
To find the enthalpy of formation of CH₄, we can reverse the combustion reaction of CH₄ and add the enthalpies of the other two combustion reactions:
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3 kJ/mol
C(graphite) + O₂(g) → CO₂(g) ΔH = -393.5 kJ/mol
2H₂(g) + O₂(g) → 2H₂O(l) ΔH = -571.6 kJ/mol
Adding these equations gives:
C(graphite) + 2H₂(g) → CH₄(g) ΔH = -74.8 kJ/mol
Therefore, the enthalpy of formation of CH₄ is -74.8 kJ/mol.
The correct answer is (i) -74.8 kJ mol-1.
6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Ans : The reaction will be possible at any temperature” is incorrect.
Here’s why:
While a reaction with a positive entropy change may be more likely to be spontaneous, other factors like enthalpy also play a role.
7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Ans : To calculate the change in internal energy (ΔU) for a process, we can use the first law of thermodynamics:
ΔU = Q – W
where:
ΔU is the change in internal energy
Q is the heat absorbed by the system
The system does work W.
Substituting the given values:
ΔU = 701 J – 394 J
ΔU = 307 J
Therefore, the change in internal energy for the process is 307 Joules.
8. The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN(g) + 3 2 O2(g) → N2(g) + CO2(g) + H2O(l)
Ans : To calculate the enthalpy change (ΔH) for the reaction, we can use the relationship between ΔH and ΔU:
ΔH = ΔU + ΔnRT
where:
ΔH is the enthalpy change
ΔU is the internal energy change
Δn is the change in the number of moles of gas
The gas constant is R.
(8.314 J/mol K)
T is the temperature in Kelvin
In this case:
ΔU = -742.7 kJ/mol
R = 8.314 J/mol K
T = 298 K
To find Δn, we need to compare the number of moles of gas on the reactant side and the product side:
Reactants: 1 mole NH2CN + 1.5 moles O2 = 2.5 moles gas
Products: 1 mole N2 + 1 mole CO2 + 1 mole H2O = 3 moles gas
Therefore, Δn = 3 – 2.5 = 0.5 moles
Substituting the values into the equation:
ΔH = -742.7 kJ/mol + (0.5 mol)(8.314 J/mol K)(298 K)
ΔH ≈ -743.9 kJ/mol
Therefore, the enthalpy change for the reaction at 298 K is approximately -743.9 kJ/mol.
9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.
Ans : The mass of aluminum (m) is 60 grams, and its molar mass is 27 grams per mole.
Therefore, the number of moles of aluminum is 60 grams / 27 grams/mol
= 2.22 moles.
The molar heat capacity (C) of aluminum is 24 joules per mole per Kelvin.
The temperature change (∆T) is 55 degrees Celsius – 35 degrees Celsius = 20 Kelvin.
The heat evolved (q) can be calculated using the formula:
q = C x m x ∆T.
Substituting the values, we get:
q = (24 J/mol K) x (2.22 mol) x (20 K)
= 1065.6 joules
= 1.067 kilojoules.
10. Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C.
Cp [H2O(l)] = 75.3 J mol–1 K–1
Cp [H2O(s)] = 36.8 J mol–1 K–1
Ans :
11. Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
Ans : To calculate the heat released upon the formation of 35.2 g of CO2, we need to first determine the number of moles of CO2 formed.
Step 1:
Find the number of moles of CO2:
The molar mass of CO2 is 12 g/mol (for carbon) + 16 g/mol (for oxygen) + 16 g/mol (for oxygen) = 44 g/mol.
Number of moles of CO2 =
mass of CO2 / molar mass of CO2
Number of moles of CO2 =
35.2 g / 44 g/mol
Number of moles of CO2 =
0.8 mol
Step 2: Calculate the heat released:
Carbon’s combustion enthalpy to CO2 is -393.5 kJ/mol. This means that for every mole of CO2 formed, 393.5 kJ of heat is released.
Heat released =
(number of moles of CO2) * (enthalpy of combustion of CO2)
Heat released =
0.8 mol * (-393.5 kJ/mol)
Heat released ≈ -314.8 kJ
Therefore, the heat released upon the formation of 35.2 g of CO2 from carbon and dioxygen gas is approximately -314.8 kJ.
12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆r H for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
Ans : The enthalpy of reaction (∆rH) can be calculated using the following equation:
∆rH = ΣnH°(products) – ΣnH°(reactants)
Where:
n = stoichiometric coefficient of each species
H° is the standard enthalpy of formation
For the given reaction, the equation becomes:
∆rH = [81 + 3(-393)] – [9.7 + 3(-110)] = -778 kJ/mol
This means that the reaction is exothermic, releasing 778 kilojoules of energy per mole of reactants consumed.
13.Given N2(g) + 3H2(g) → 2NH3(g); ∆r H = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas?
Ans : The standard enthalpy of formation of NH3 gas can be calculated using the given reaction and its enthalpy change.
The balanced chemical equation for the formation of NH3 gas is:
N2(g) + 3H2(g) → 2NH3(g)
The enthalpy change for this reaction is -92.4 kJ/mol.
ΔHf represents the amount of heat absorbed or released when one mole of a compound is formed from its constituent elements under standard conditions. In this case, the standard enthalpy of formation of NH3 gas is half of the enthalpy change for the given reaction, since the reaction produces 2 moles of NH3.
Hence, the standard enthalpy of formation of NH3 gas is:
ΔHf(NH3) = -92.4 kJ/mol / 2 = -46.2 kJ/mol
NH3 has a standard enthalpy of formation of -46.2 kJ/mol.
14. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) + 3 2 O2(g) → CO2(g) + 2H2O(l) ; ∆r H = –726 kJ mol–1
C(graphite) + O2(g) → CO2(g) ; ∆c H = –393 kJ mol–1
H2(g) + 1 2 O2(g) → H2O(l); ∆f H = –286 kJ mol–1.
Ans : The given equations represent a Hess’s law cycle to calculate the standard enthalpy of formation (∆fH°) of methanol (CH₃OH).
The target equation is:
C(s) + 2H₂(g) + 1/2O₂(g) → CH₃OH(l)
To obtain this equation, the following steps are performed:
Multiply equation (iii) by two and sum it with equation (ii).
This gives:
C(s) + 2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
∆H = -965 kJ/mol
Subtract equation (iv) from equation (i):
This gives:
CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l)
∆H = -726 kJ/mol
Subtract the resulting equation from the equation obtained in step 1:
This gives the target equation:
C(s) + 2H₂(g) + 1/2O₂(g) → CH₃OH(l)
∆fH° = -239 kJ/mol
Therefore, the standard enthalpy of formation of methanol is -239 kJ/mol. This means that the formation of one mole of methanol from its elements in their standard states releases 239 kJ of energy.
15. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g)
and calculate bond enthalpy of C – Cl in CCl4(g).
∆vapH(CCl4) = 30.5 kJ mol–1.
∆f H (CCl4) = –135.5 kJ mol–1.
∆aH (C) = 715.0 kJ mol–1, where ∆aH is enthalpy of atomisation
∆aH (Cl2) = 242 kJ mol–1
Ans :
16. For an isolated system, ∆U = 0, what will be ∆S ?
Ans : For an isolated system, if ΔU = 0, then ΔS must be greater than or equal to 0.
This is a consequence of the Second Law of Thermodynamics, which states that the total entropy of an isolated system always increases or remains constant in any spontaneous process.
If ΔU = 0 (meaning there is no change in internal energy), then the only way for the system’s entropy to change is through heat transfer with the surroundings. However, since the system is isolated, there can be no heat transfer. Therefore, the entropy of the system must remain constant or increase.
In summary:
ΔU = 0 for an isolated system means there is no change in internal energy.
ΔS ≥ 0 for an isolated system, according to the Second Law of Thermodynamics.
If ΔU = 0, then ΔS must be greater than or equal to 0.
17. For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.
Ans : The spontaneity of a reaction is determined by the Gibbs free energy change (ΔG), which is related to the enthalpy change (ΔH) and entropy change (ΔS) by the equation:
ΔG = ΔH – TΔS
where:
ΔG is the Gibbs free energy change (in J/mol)
ΔH is the enthalpy change (in J/mol)
T is the temperature in Kelvin
ΔS is the entropy change (in J/mol K)
A reaction is spontaneous if ΔG is negative
Given:
ΔH = 400 kJ/mol = 400000 J/mol
ΔS = 0.2 kJ/mol K = 200 J/mol K
T = 298 K
Substituting the values:
ΔG = 400000 J/mol – (298 K)(200 J/mol K)
ΔG ≈ 340400 J/mol
Since ΔG is positive at 298 K, the reaction is non-spontaneous at this temperature.
To find the temperature at which the reaction becomes spontaneous, we need to set ΔG equal to zero and solve for T:
0 = 400000 J/mol – T(200 J/mol K)
T = 2000 K
Hence, the reaction will become spontaneous at a temperature of 2000 K.
18.For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?
Ans : For the reaction 2Cl(g) → Cl2(g), the signs of ΔH and ΔS are as follows:
ΔH is negative.
The reaction gives off energy to the environment in the form of heat.. This is because a bond is being formed between the two chlorine atoms, which releases energy.
ΔS is negative. This means the reaction decreases entropy. This is because two individual chlorine atoms are combining to form one molecule, which results in a decrease in the randomness or disorder of the system
19. For the reaction 2 A(g) + B(g) → 2D(g) ∆U = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously.
Ans : The given thermodynamic data includes the change in internal energy (ΔU), the change in the number of moles of gas (Δng), the gas constant (R), and the temperature (T).
Using the equation ΔH° = ΔU° + ΔngRT, we can calculate the enthalpy change (ΔH°) for the reaction.
Substituting the given values, we get:
ΔH° = (-10.5 kJ) + (-1 mol) x (8.314 J/mol K) x (298 K)
= -12.978 kJ
According to the Gibbs-Helmholtz equation, ΔG° = ΔH° – TΔS°, where ΔG° is the standard Gibbs free energy change, ΔS° is the standard entropy change, and T is the temperature.
Substituting the given values, we get:
ΔG° = (-12.978 kJ) – (298 K) x (-0.0441 kJ/K)
= 0.164 kJ
Therefore, the standard Gibbs free energy change for the reaction is 0.164 kJ.
The positive value of ΔG° indicates that the reaction is nonspontaneous.
20. The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K.
Ans : The relationship between the equilibrium constant (K) and the standard free energy change (ΔG°) is given by the equation:
ΔG° = -RT ln K
where:
ΔG° =
standard free energy change (in J/mol)
R =
gas constant (8.314 J/mol K)
T =
temperature in Kelvin (300 K in this case)
K =
equilibrium constant (10 in this case)
Putting the values:
ΔG° =
-(8.314 J/mol K) * (300 K) * ln(10)
ΔG° ≈ -5744 J/mol
Converting to kilojoules:
ΔG° ≈ -5.74 kJ/mol
Hence, the value of ΔG° for the reaction is approximately -5.74 kJ/mol.
The negative sign indicates that the reaction is spontaneous under standard conditions.
21. Comment on the thermodynamic stability of NO(g), given
1 2 N2(g) + 1 2 O2(g) → NO(g); ∆r H = 90 kJ mol–1
NO(g) + 1 2 O2(g) → NO2(g): ∆r H= –74 kJ mol–1
Ans : To analyze the thermodynamic stability of NO(g), we need to consider the enthalpy changes (ΔrH) for the given reactions.
Reaction 1:
1/2 N2(g) + 1/2 O2(g) → NO(g)
ΔrH = 90 kJ/mol
The positive enthalpy change indicates an endothermic reaction, absorbing heat.”
Reaction 2:
NO(g) + 1/2 O2(g) → NO2(g)
ΔrH = -74 kJ/mol
The negative enthalpy change indicates that the reaction is exothermic, releasing energy to the environment.
Thermodynamic Stability:
A compound is considered thermodynamically stable if it tends to remain in its current state under given conditions. In other words, a stable compound is one that is not prone to spontaneous decomposition or reaction with other substances.
Based on the given reactions:
NO(g) is not thermodynamically stable. This is because it can react with O2(g) to form NO2(g), releasing energy in the process. This suggests that NO(g) is a relatively high-energy compound that is prone to further reactions.
NO2(g) is thermodynamically more stable than NO(g) because its formation from NO(g) and O2(g) releases energy.
Therefore, while NO(g) can exist under certain conditions, it is not considered a highly stable compound due to its tendency to react with oxygen to form NO2(g).
22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1.
Ans : The heat transferred to the surroundings (qrev) is equal to the negative of the enthalpy change (ΔH°),
which is -286 kJ/mol or 286000 J/mol.
The entropy change of the surroundings (ΔSsurroundings) is calculated by dividing the heat transferred by the temperature (T).
Substituting the values, we get:
ΔSsurroundings = qrev / T = (286000 J/mol) / (298 K)
= 959 J/K mol.
Therefore, the entropy change of the surroundings is 959 J/K mol.
