Question:

Change in enthalpy for reaction 2H2O2(l) → 2H2O(l) + O2(g) If heat of formation of H2O2(l) and H2O(l) are – 188 & – 286 KJ/mol respectively : -

Updated On: Sep 23, 2024
  • -196 KJ/mol
  • (+196 KJ/mol)
  • (+198 KJ/mol)
  • -198 KJ/mol
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The Correct Option is A

Solution and Explanation

The correct option is(A): -196 KJ/mol. The change in enthalpy for the given reaction is calculated using the heats of formation of the reactants and products: ΔH = Σ(heats of formation of products) - Σ(heats of formation of reactants) ΔH = [2(-286 KJ/mol) + 0] - [2(-188 KJ/mol)] = -572 KJ/mol + 376 KJ/mol = -196 KJ/mol So, the change in enthalpy for the reaction 2H2O2(l) → 2H2O(l) + O2(g) is -196 KJ/mol.
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Concepts Used:

Enthalpy change

Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.

Formula for change in enthalpy is:-

dH = dU + d(PV)

The above equation can be written in the terms of initial and final states of the system which is defined below:

UF – UI = qP –p(VF – VI)

Or qP = (UF + pVF) – (UI + pVI)

Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained: 

qP = HF – HI = ∆H

Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,

∆H = ∆U + p∆V

Standard Enthalpy of Reaction

To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.