Chemical Thermodynamics Important Questions

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Very Short Answer Questions (1 mark)

Ques. How many laws of thermodynamics are present?

Ans. There are four laws of thermodynamics.

Ques. What is surrounding?

Ans. Everything that lies outside the boundary of a system that affects the behaviour of the system is called surroundings.

Ques. Name the variables of the state of the system?

Ans. Pressure, Volume, temperature and amount are variables of the state of the system.

Ques. State the equation when a change of state is calculated by transfer of heat?

Ans. The equation when change of state is calculated by transfer of heat is as follows:

Δu = q + w

Ques. It is calculated that one mole of propanone requires less heat to evaporate than 1 mole of water. Which liquid has a higher enthalpy of vaporization?

Ans. Propanone requires less amount of heat for vaporization due to the presence of a weak force of attraction between molecules. Thus, it can be said that water has a higher enthalpy of vaporization.

Ques. In what type of system is coffee kept in a cup?

Ans. Coffee is kept inside an open system as coffee can exchange water vapour and energy with the surroundings.

Ques. Define enthalpy?

Ans. Enthalpy is defined as the total heat that is found in the system.

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Short Answer Question (2 marks)

Ques. State the zeroth law of thermodynamics?

Ans. The Zeroth law of thermodynamics states that if two bodies are placed in a system that is in thermal equilibrium with respect to the third body, then the first and second bodies are in a state of thermal equilibrium.

Ques. State third law of thermodynamics?

Ans. The third law of thermodynamics states that when the temperature of the system reaches the state of absolute zero, then the required entropy of the system will become constant.

Ques. Explain the relationship between free expansion of gas and isothermal?

Ans. The following derivation can explain the relationship between the free expansion of gas and isothermal. In this case, the value of Pex is equal to 0.

• As a result of which 

Δ U = 0, q = 0 

• When isothermal irreversible change occurs, then

q = – W = P ex (Vf – Vi)

• When isothermal reversible change occurs in such cases:

q = – W = nRT log V₁/V₂

2.303 nRT log V₁/V₂

Ques. Consider the enthalpy of atomisation for the reaction CH₃(g)→ C(g) + 3H (g) is 1660 kJ mol^–1. What is the bond energy of the C–H bond?

Ans. Enthalpy of atomisation of 3 moles of C−H bonds = 1660 kJ mol^–1

• C−H bond energy, per mole = 1660 kJ mol^–1/3 = 533.33 kJmol^–1.

Ques. Explain the first law of thermodynamics.

Ans. The first law of thermodynamics states that energy can neither be created nor it can be destroyed. The energy is transformed from one form to another, which goes by the equation

ΔQ = ΔU + ΔW.

where

• ΔQ is the heat given to a thermodynamic system
• ΔW is the Work Done
• ΔU is the Internal Energy

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Long Answer Question (3 marks)

Ques. Determine the factors which are changed when internal energy is created?

Ans. The factors which are changed when internal energy is created are as follows:

• The work which is performed by the system.
• Heat created by the system moves inside or outside of the system.
• The matter will leave or enter the system.

Ques. Explain the conventions of thermochemical equations?

Ans. The conventions of thermochemical equations are as follows:

• The coefficients used in the thermochemical equation determine the number of moles used in the reactants and products.
• If the value of the chemical reaction is reversed then in such case the value of Δt HΘ is reversed in sign. 
• The value of Δt HΘ is defined as the number of moles of equation that is used by an equation.

Ques. Explain Gibbs energy function?

Ans. Gibbs energy function also known as Gibbs free energy is defined as the maximum amount of work that is performed by a closed system.

• In gibbs free energy the value of temperature and pressure is constant.
• It can be represented by the equation: ΔG° = ΔH° – TΔS°. 
• Where, H is the enthalpy, S is the entropy and T is the temperature.
• If the value of ΔG is positive, it means that the process is nonspontaneous. 
• If the value of ΔG is negative, it means that the process is spontaneous. 

Ques. Enthalpy of combustion of carbon to carbon dioxide is -390.5 KJ mol -1. Calculate the heat released upon formation of 65.5g of CO2 from carbon and oxygen gas. 

Ans. C(s) + O₂(g) → CO₂(g); ΔH = -390.5 KJ mol ^-1

• Heat released in the formation of 44g of CO₂ = -390.5 KJ
• Heat released in the formation of 65.5g of CO₂ = (390.5 KJ) x (65.5g)/(44g) = 581.31KJ

Ques. Calculate ΔG at 280 K for the reaction,
2NO + O₂ → 2NO₂ when ΔH and ΔS of the reaction are -700 J and -0.35 J/K respectively.

Ans. Given, ΔH = -700 J, ΔS = -0.25 J/K and T = 280 K

• We know that,
• ΔG = ΔH – TΔS
• ΔG = (-700) -280(0.35)
• ΔG = -700 – 98
• ΔG = -798 J

Ques. Suppose the standard molar entropy of H₂O (l) is 80 J K^–1 mol^–1. Will the standard molar entropy of H₂O (s) be more, or less than 80 J K^–1 mol^–1?

Ans. It is known that the solid form of H₂O is ice. When we talk about ice, molecules of ice are less random than in water.

• So, molar entropy of H₂O (s) < molar entropy of H₂O (l). The standard molar entropy of
• H₂O (s) is less than 80 J K^–1 mol^–1

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Very Long Answer Question (5 marks)

Ques. If an ideal heat engine operates in a Carnot cycle between 1200 K and 1000 K and if it absorbs 6000 J of heat at a higher temperature then find the heat supplied from the source.

Ans. Given, T₁ = 1200 K and T₂ = 1000 K

• Heat Absorbed at High Temperature = 6000 J
• Heat Supplied from Source =?
• Efficiency of Heat Engine (E) = 1 – (T₂ / T₁)
• Efficiency of Heat Engine (E) = 1 – 1000/1200 = 1- 5/6
  Efficiency of Heat Engine (E) = 1/6
• We know that, E = Heat Supplied from Source/Heat Absorbed at High Temperature
• 1/6 = Heat Supplied from Source/6000
• Heat Supplied from Source = 1000 J

Ques. What amount of heat must be supplied to 4.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N₂ = 28, R = 8.3 J K^–1 mol^–1)

Ans. Here, mass of gas , m = 4 x 10^-2 kg = 40 g

• Rise in temperature, Δ T = 45°C
• Heat required, Δ Q = ?; Molecular mass, M = 28
• Number of moles, n= m/M = 40/28 = 1.42
• As nitrogen is a diatomic gas, molar specific heat at constant pressure is 
• Cp = 7/2R=7/2 x 8.3 J K^–1 mol^–1
• as Δ Q = nCpΔ T
• Δ Q = 1.42 x 7/2 x 8.3 x 45 J = 1856.295J

Ques. A geyser heats water flowing at a rate of 3.0 litre per minute from 37°C to 87°C. If the geyser operates on a gas burner, find out the rate of the consumption of fuel if its heat of combustion is 2.0 x 10⁴ J/g. 

Ans. The volume of water heated is 3.0 litre per min
Mass of water heated is m = 4000 g per min 

• Increase in temperature,
• Δ T = 87°C – 37°C = 50°C
• Specific heat of water, c = 4.2 Jg ^-1 °C^-1
• amount of heat used, Q = mc Δ T
• or Q = 4000 g min ^-1 x 4.2 Jg ^-1 °C^-1 x 50°C
• 84 x 10⁴ J min^-1
• Rate of combustion of fuel = 84 x 10⁴ / 2.0 x 10⁴ = 21g min^-1

Ques. Calculate the energy in Joule that must be given to freeze two kg of water, assuming the residential refrigerator is a reversible engine operating between the melting point of ice and the room temperature of 37°C. L = 80 cal g^-1 when the water temperature is 0°C.

Ans. T₁ = 37 + 273 = 310 K in this case.

• T₂ = 0 + 273 = 273 K
• m = 2 kg = 1000 g
• L = 80 cal g^-1
• Q₂ = mL = 1000
• 80 cal = 8 × 10⁴ cal to be eliminated heat
• Using the relation
• Q₁ / Q₂ = T₁ / T₂, we get
• Q₁ = T₁ / T₂ × Q₂ = 310 / 273 × 8 × 10⁴
• = 90842.1 cal
• Energy required to be supplied,
• W = Q₁ – Q₂ = (90842.1 – 80000) cal
• = 10842.1 cal
• = 10842.1 × 4.2 J = 45536.82 J

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