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The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) is
  • a)
    - 333.14 kJ mol-1
  • b)
    333.14 kJ mol-1
  • c)
    - 228.54 kJ mol-1
  • d)
    228.54 kJ mol-1
Correct answer is option 'C'. Can you explain this answer?
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The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 k...
Given data:
Enthalpy change of reaction, ΔH = -57.3 kJ mol-1
Enthalpies of formation, ΔHf(H (aq)) = 0 kJ mol-1 and ΔHf(H2O(l)) = -285.84 kJ mol-1

Enthalpy of formation of OH- can be calculated using Hess's Law, which states that the enthalpy change of a reaction is the same whether it occurs in one step or in a series of steps.

Steps to calculate the enthalpy of formation of OH-:
1. Write the balanced chemical equation for the reaction.
H (aq) + OH- (aq) → H2O (l)
2. Write the enthalpy change for the given reaction.
ΔH = -57.3 kJ mol-1
3. Write the enthalpy change for the reverse reaction.
H2O (l) → H (aq) + OH- (aq)
ΔH = +57.3 kJ mol-1
4. Write the enthalpy change for the formation of H2O (l) from H (aq) and OH- (aq), which is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
ΔHf(H2O(l)) = ΣΔHf(products) - ΣΔHf(reactants)
ΔHf(H2O(l)) = 0 - 285.84 kJ mol-1 = -285.84 kJ mol-1
5. Write the enthalpy change for the formation of OH- (aq) from H (aq) and H2O (l), which is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
ΔHf(OH-(aq)) = ΣΔHf(products) - ΣΔHf(reactants)
ΔHf(OH-(aq)) = -285.84 kJ mol-1 - (+57.3 kJ mol-1) = -228.54 kJ mol-1

Therefore, the enthalpy of formation of OH- (aq) is -228.54 kJ mol-1, which is option (c).
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The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) isa)- 333.14 kJ mol-1b)333.14 kJ mol-1c)- 228.54 kJ mol-1d)228.54 kJ mol-1Correct answer is option 'C'. Can you explain this answer?
Question Description
The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) isa)- 333.14 kJ mol-1b)333.14 kJ mol-1c)- 228.54 kJ mol-1d)228.54 kJ mol-1Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) isa)- 333.14 kJ mol-1b)333.14 kJ mol-1c)- 228.54 kJ mol-1d)228.54 kJ mol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) isa)- 333.14 kJ mol-1b)333.14 kJ mol-1c)- 228.54 kJ mol-1d)228.54 kJ mol-1Correct answer is option 'C'. Can you explain this answer?.
Solutions for The enthalpy change of the reaction H+(aq) + OH-(aq) H2O(l) is -57.3 kJ mol-1. If the enthalpies of formation of H+(aq) and H2O(l) are zero and -285.84 kJ mol-1, respectively, then the enthalpy of formation of OH-(aq) isa)- 333.14 kJ mol-1b)333.14 kJ mol-1c)- 228.54 kJ mol-1d)228.54 kJ mol-1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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