Enthalpy is a thermodynamic parameter that gauges the total heat present in a thermodynamic system where the pressure is regular. It is equivalent to the sum of the internal energy and the product of the pressure and volume of a thermodynamic system.
H = E + PV
Various factors that affect the enthalpy of an atom are the amount of reactant and product, the physical state of reactants and products, allotropic modification, temperature and pressure.
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Definition: The enthalpy change of an atom is defined as the amount of heat absorbed or evolved in a reaction carried out at a steady pressure. |
Enthalpy Change Chemistry Questions with Solutions
Q1. If the activation energy is equal for both forward and backward reactions, then
(a) Δ H = 0
(b) Δ G = 0
(c ) Δ S = 0
(d) Δ H = Δ G = Δ S = 0
Answer: (a) If the activation energy is equal for both forward and backward reactions, then Δ H should be zero.
Q2. The change in enthalpy of a system is equivalent to the heat absorbed by the system at a
(a) Constant temperature
(b) Constant pressure
(c ) Constant volume
(d) None of the above
Answer: (b) The change in enthalpy of a system is equivalent to the heat absorbed by the system at a constant pressure.
Q3. The change in enthalpy of an exothermic reaction is
(a) Always positive
(b) Always negative
(c ) Can either be positive or negative
(d) None of the above
Answer: (b) The change in enthalpy of an exothermic reaction is always negative.
Q4. If the heat is transmitted to a system at a steady pressure. In that case, the enthalpy of the system will
(a) Increase
(b) Decrease
(c ) First increase then decrease
(d) First decrease then increase
Answer: (a) If the heat is transmitted to a system at a steady pressure, the enthalpy of the system will increase.
Q5. The change in enthalpy when 1 mole of the compound is formed under standard conditions is known as
(a) Standard enthalpy of neutralisation
(b) Standard enthalpy of formation
(c ) Standard enthalpy of combustion
(d) None of the above
Answer: (b) The change in enthalpy when 1 mole of the compound is formed under standard conditions is known as standard enthalpy of formation.
Q6. What is enthalpy?
Answer: Enthalpy is a thermodynamic parameter that gauges the total heat present in a thermodynamic system where the pressure is regular. It is equivalent to the sum of the internal energy and the product of the pressure and volume of a thermodynamic system.
H = E + PV
Q7. What are the various factors that affect the enthalpy of an atom?
Answer: Enthalpy is a thermodynamic parameter that gauges the total heat present in a thermodynamic system where the pressure is regular. Various factors that affect the enthalpy of an atom are mentioned below.
1. Amount of reactant and product.
2. Physical State of Reactants and Products.
3. Allotropic Modification
4. Temperature and Pressure
Q8. What is the enthalpy change?
Answer: The enthalpy change of an atom is defined as the amount of heat absorbed or evolved in a reaction carried out at a steady pressure.
Q9. Calculate the enthalpy change for the following reaction:
CH4 (g) + 2 O2 (g) ⟶ CO2 (g) + 2 H2O (l)
Given that enthalpies of formation of CH4, CO2 and H2O are 74.8 kJ mol−1, − 393.5 kJ mol−1, and − 286 kJ mol−1, respectively.
Answer: Enthalpy of formation of CH4 = 74.8 kJ mol−1,
Enthalpy of formation of CO2 = − 393.5 kJ mol−1
Enthalpy of formation of H2O = − 286 kJ mol−1
Enthalpy change = Δ HoPRODUCTS – Δ HoREATANTS
Enthalpy change = (Δ HoCO2 + Δ HoH2O) – (Δ HoCH4 + 2 Δ HoO2)
Enthalpy change = (− 393.5 kJ mol−1 + − 286 kJ mol−1) – (74.8 kJ mol−1 + 2 x 0)
Enthalpy change = – 890.7 kJ mol−1
Q10. Calculate the enthalpy of formation of OH– ions at 25o C from the following thermochemical data.
H2O (l) → H+ (aq) + OH– (aq); Δ HO = 57.3 kJ
H2 + ½ O2 (g) → H2O (l); Δ HO = – 285.9 kJ
Answer: Foremost, we will add first two-equation.
H2 (g) + ½ O2 (g) → H+ (aq) + OH– (aq); Δ HO = 57.3 kJ – 285.9 kJ = – 228.6 kJ.
Hence, Δ HO = – 228.6 kJ = 0 + Δ HfO (OH– (aq)) – (0+0), since, by convention, Δ HfO [H+(aq)] = 0,
Hence, Δ HfO [OH– (aq)] = – 228.6 kJ.
Q11. The enthalpy of combustion of glucose C6H12O6 (s) is – 2816 kJ mol-1 at 25o C. Calculate Δ HfO C6H12O6. The HfO values for CO2 (g) and H2O (l) are – 393.5 and – 285.9 kJ mol-1, respectively.
Answer: C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l); Δ HO = – 2816 kJ.
Since Δ H = ∑ Δ HfO (products) – ∑ Δ HfO (reactants), we find that
– 2816 kJ = (6 X 393.5 kJ mol-1) + (6 X – 285.9 kJ mol-1) – Δ HfO (C6H12O6) – 6 Δ HfO (O2)
We know that, Δ HfO (O2) = 0
So, Δ HfO (C6H12O6) = – 1260.4 kJ mol-1.
Q12. Calculate the enthalpy of combustion of methane at 25o C and 1 atm pressure.
Given that Δ HfO (CO2) = – 393.5 kJ mol-1, Δ HfO (H2O) = – 285.9 kJ mol-1 and Δ HfO (CH4) = -74.8 kJ mol-1.
Answer: The combustion of methane is referenced as
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O
Δ Ho = Δ Hfo (CO2) + 2 Δ Hfo (H2O) – Δ Hfo (CH4) – 0
Δ Ho = (- 393.5 kJ mol-1 ) + 2 X (- 285.9 kJ mol-1) – (- 74.8 kJ mol-1)
Δ Ho = – 890.5 kJ mol-1.
Thus, the enthalpy of combustion of methane at 25o C and 1 atm pressure = – 890.5 kJ mol-1.
Q13. One mole of a non-ideal gas undergoes a state change from (2 atm, 3 L, 95 K) to (4 atm, 5 L, 245 K) with a change of internal energy, Δ U = 30 L atm. What is the difference in enthalpy (Δ H)?
Answer: Change in enthalpy = Δ H = Δ U + Δ (PV)
Change in enthalpy = Δ H = Δ U + (P2 V2 – P1 V1)
Change in enthalpy = Δ H = 30 + (20 – 6)
Change in enthalpy = Δ H = 44 L atm.
Q14. 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.
NH2CN (g) + 23 O2 (g) → N2 (g) + CO2 (g) + H2O (l)
Calculate the enthalpy change for the reaction at 298 K?
Answer: For the given reaction, Δ n = 1 + 1 − 1.5 = 0.5.
Moreover, Δ H = Δ U + Δng RT
Δ H = − 742.7 + 0.5 X 8.314 X 10−3 X 298
Δ H = − 742.7 + 1.2
Δ H = − 741.5 kJ mole−1.
Q15. The combustion of one mole of benzene occurs at 298 K, and 1 atm after combustion, CO2 (g) and H2O (l) are produced, and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formulation Δ Hf of benzene. Given the standard enthalpy of formation of CO2 (g) and H2O (l) are – 393.5 KJ mole-1 and – 285.83 KJ mole-1.
Answer: Reaction: C6H6 + 15/2 O2 → 6 CO2 + 3 H2O, Δ Hrxn = − 3267 KJKJ mole-1
Δ Hrxn = 6 Δ Hf CO2 + 3 Δ Hf H2O − Δ Hf C6H6
Δ Hf (Benzene) = 6 X (− 393.5) + 3 X (− 285.8) + 3267
Δ Hf (Benzene) = − 3218.49 + 3267
Δ Hf (Benzene) = 48.51 KJ.
Practise Questions on Enthalpy Change
Q1. What is Hess law? Explain the feasibility of Hess law with an example.
Q2. Calculate the standard enthalpy of formation of methanol using the following data.
C (graphite) + O2 (g) → CO2 (g) (ΔH = − 393 kJmol−1)
CH3OH (l) + 2 O2 (g) → CO2 (g) + 2 H2O (l) (ΔH = − 726 kJmpl−1)
H2 (g) + 2 O2 (g) → H2O (l) (ΔH = − 286 kJmol−1)
Q3. How will you differentiate between enthalpy from entropy?
Q4. How will you differentiate between extensive and intensive functions?
Q5. If the combustion of 1 g of graphite produces 20.7 kJ of heat. What will be the molar enthalpy change? What is the significance of the sign of enthalpy?
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