Solved Practice Questions on Internal Energy, Work Done and Enthalpy | Physical Chemistry PDF Download

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Q.1. Why we cannot determine the exact or absolute value of internal energy?

It is impossible to determine an object's internal energy in its exact or absolute form. This is because most of the quantities that contribute to the system's internal energy are impossible to estimate with accuracy.

Q.2. Calculate the internal energy change in each of the following cases: -
(i) A system absorbs 25 kJ of heat and does 15 kJ of work.
(ii) 15 kJ of work is done on the system and 25 kJ of heat is given out by the system.

(i) Here, q= +25 kJ and w= -15 kJ
According to the first law of thermodynamics,
ΔE = q + w = +25 -15 = 10 kJ
This implies the internal energy of the system increases by 10 kJ.
(ii) Here, q= -25 kJ and w= +15 kJ
According to the first law of thermodynamics,
ΔE = q + w = -25 +15 = -10 kJ
This implies the internal energy of the system decreases by 10 kJ.

Q.3. A chemist while studying the properties of gaseous CCI₂F₂, a chlorofluorocarbon refrigerant cooled a 1.25 g sample at constant atmospheric pressure of 1.0 atm from 320 K to 293 K. During cooling, the sample volume decreased from 274 to 248 mL. Calculate ΔE for the chlorofluorocarbon for this process. For CCI₂F₂, C_p = 80.7 J/(mol K).

ΔH = q_p and C_p is the heat evolved or absorbed per mole of 1o fall or rise in temperature.
Here, fall in temperature = 320-293 = 27 K.
Molar mass of CCI₂F₂ = 121 g/mol
Therefore, heat evolved from 1.25 g of the sample on being cooled from 320 K to 293 K at constant pressure =( 80.7/121) × 1.25 × 27 J = 22.51 J.
Further ΔH = ΔU + PΔV = -22.51 J
PΔV = 1 atm × [(248-274)/1000] L = -0.026 L atm = -0.026 × 101.35 J = -2.63 J
Therefore, -22.51 = ΔU + (-2.63) ΔU = -19.88 J

Q.4. Calculate the standard internal energy change for the reaction OF₂ (g) + H₂O (g)→ O₂ (g) + 2HF (g) at 298 K, given that the enthalpies of formation of OF₂(g), H₂O(g) and HF(g) are +20, -250 and -270 kJ/mol respectively.

Step I. Calculation of standard enthalpy of reaction.
ΔH_f^o = ∑ΔH_f^o(products) - ∑ΔH_f^o(reactants)
ΔH_f^o = [ΔH_f^o(O₂) + 2ΔH_f^o(HF)] - [ΔH_f^o(OF₂) + ΔH_f^o(H₂O)]
ΔH_f^o = [0 + 2(-270)] - [+20 + (-250)] kJ/mol
ΔH_f^o = -310 kJ/mol
Step II. Calculation of standard internal energy change.
For the given reaction,
Δn_g = (1+2) - (1+1) = 1
ΔH^o = ΔU^o + Δn_gRT
Or ΔU^o = ΔH^o - Δn_gRT
ΔU^o = -310 - [(1) × (8.314× 10^-3 kJK^-1mol^-1) × (298 K)]
ΔUo = -312.48 kJmol^-1

Q.5. For the reaction of one mole of HCl with zinc dust in a bomb calorimeter, how does ΔU and ΔW corresponds?

The bomb calorimeter, which consists of a sealed combustion chamber known as a bomb, is frequently used to measure the heat of combustion of organic substances. No expansion or compression is allowed during a procedure if the container is sealed. so W = 0 and ΔU = q.

Q.6. For the reaction X₂O₄ (l) → 2XO₂ (g) at 298 K, given the values, ΔU = 9 kJ and ΔS = 84 JK-1. Calculate ΔG.

Given reaction is X₂O₄ (l) → 2XO₂ (g), T= 298 K
ΔU = 9 kJ and ΔS = 84 JK-1, ΔH = ?
Δng = 2-0 = 2
ΔH = ΔU+ Δn_gRT = 9kJ + 2×8.314×298 = (9000 J + 4955.144) J = 13955.144 J
ΔG = ΔH - TΔS = 13955.144 J - (298 K × 84 JK^-1) = -11076.856 J = -11.08 kJ