Example Problems on Hess Law

Problem 1: Calculate the reaction’s standard enthalpy change using the following reaction.

CO_2(g) + H_2(g) → CO_(g) + H₂O_(g)

Given that, Δ_rH^o for CO_(g), CO_2(g), and H₂O_(g) as -110.5 kJ/mol, -393.5 kJ/mol, and 241.8kJ/mol respectively.

Solution:

Δ_rH^o for the reaction can be given as:

△_rH^o = Σ△_fH^o_(Products) – Σ△_fH^o_(Reactants)

Given:

• Δ_f​H^∘(CO(g)) = −110.5kJ/mol
• Δ_f​H^∘(CO₂​(g)) = −393.5kJ/mol
• Δ_f​H^∘(H₂​O(g)) = 241.8kJ/mol
• We know, △_fH^o (H₂) = 0

Thus, △_rH^o= [△_fH^o (H₂O) + △_fHo(CO)] – [△_fH^o (CO₂) + △_fH^o (H₂)]

⇒ △_rH^o=[−110.5+241.8]−[−393.5+0]

⇒ △_rH^o= (131.3) − (−393.5)

⇒ △_rH^o= 131.3 + 393.5

⇒ △_rH^o= 524.8kJ/mol

Problem 2: Hess’s law states that a chemical reaction is independent of the route of chemical reactions while keeping the same

(a) Initial and Final Conditions

(b) Initial conditions only

(c ) Final conditions only

(d) None of the above

Solution:

Hess’s law states that a chemical reaction is independent of the route of chemical reactions while keeping the same initial and final conditions.

Problem 3: Calculate the enthalpy change for the following reaction:

CH₄ ​_(g) + 2 O_{2 ​(g)} ⟶ CO_{2 ​(g)} + 2 H₂​O _(l).

Given that enthalpies of formation of CH₄, CO₂ and H₂​O are 74.8 kJmol⁻¹,− 393.5 kJ mol⁻¹, and − 286 kJmol⁻¹, respectively.

Solution:

We know, ΔH^o = ΔH^o_(products) ​− ΔH^o_(reactants)​

⇒ ΔH^o = [ΔH^o(CO₂) ​+ 2 x ΔH^o(H₂O)] − [ΔH^o(CH₄) + 2 x ΔH^o(O₂)​]

⇒ ΔH^o = [− 393.5 + 2 X (−286.2)] − [74.8 + 2 X 0]

⇒ ΔH^o = − 393.5 – 572.4 – 74.8

⇒ ΔH^o = − 1040.7 kJ

Hess’s Law is a fundamental principle in thermodynamics that states that the total enthalpy change for a chemical reaction is independent of the pathway taken to reach the final state.

Hess’s Law is based on the first law of thermodynamics, which states that energy cannot be created or destroyed but can be converted from one form to another. Hess Law is also called Hess Law of Constant Heat. In this article, we will see Hess Law, its forms, applications, etc.