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Enthalpy of neutralization of H3PO3 acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisation of H3PO3 into its ions. (in KJ)
Correct answer is '5'. Can you explain this answer?
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Verified Answer
Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using...
H3PO3 → 2H+ + HPO3–2
2H+ + 2OH– → 2H2O
ΔH = –55.84 × 2 = –116.68
Now
–106.68 = ΔHion – 55.84 × 2
ΔHion = 5KJ/mol
ΔH = –55.84 × 2 = –116.68
Now
–106.68 = ΔHion – 55.84 × 2
ΔHion = 5KJ/mol
Most Upvoted Answer
Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using...
Understanding Enthalpy of Neutralization
The enthalpy of neutralization is the heat change when an acid reacts with a base to form water and a salt. In this case, we are dealing with the neutralization of phosphorous acid (H₃PO₃) and hydrochloric acid (HCl) by sodium hydroxide (NaOH).
Given Data
- Enthalpy of neutralization for H₃PO₃: ΔH₃PO₃ = -106.68 kJ/mole
- Enthalpy of neutralization for HCl: ΔHCl = -55.84 kJ/mole
Reaction Overview
1. Acid Dissociation:
- H₃PO₃ ionizes to produce H⁺ ions and H₂PO₃⁻ ions.
- The ionization reaction can be represented as:
H₃PO₃ → H⁺ + H₂PO₃⁻
2. Neutralization Reaction:
- The neutralization reactions can be represented as:
H₃PO₃ + NaOH → NaH₂PO₃ + H₂O
HCl + NaOH → NaCl + H₂O
Calculating ΔH ionization for H₃PO₃
The enthalpy of ionization of H₃PO₃ (ΔH_ionization) can be derived using Hess's Law. The total heat change for the dissociation of H₃PO₃ can be expressed as:
ΔH₃PO₃ = ΔH_ionization + ΔHCl
By rearranging the formula, we get:
ΔH_ionization = ΔH₃PO₃ - ΔHCl
Substituting the Values
- ΔH_ionization = (-106.68 kJ) - (-55.84 kJ)
- ΔH_ionization = -106.68 + 55.84
- ΔH_ionization = -50.84 kJ
Since we need the positive value for ionization, we take the absolute value:
Final Result
- ΔH_ionization = 5 kJ
Thus, the enthalpy of ionization of H₃PO₃ into its ions is approximately 5 kJ.
The enthalpy of neutralization is the heat change when an acid reacts with a base to form water and a salt. In this case, we are dealing with the neutralization of phosphorous acid (H₃PO₃) and hydrochloric acid (HCl) by sodium hydroxide (NaOH).
Given Data
- Enthalpy of neutralization for H₃PO₃: ΔH₃PO₃ = -106.68 kJ/mole
- Enthalpy of neutralization for HCl: ΔHCl = -55.84 kJ/mole
Reaction Overview
1. Acid Dissociation:
- H₃PO₃ ionizes to produce H⁺ ions and H₂PO₃⁻ ions.
- The ionization reaction can be represented as:
H₃PO₃ → H⁺ + H₂PO₃⁻
2. Neutralization Reaction:
- The neutralization reactions can be represented as:
H₃PO₃ + NaOH → NaH₂PO₃ + H₂O
HCl + NaOH → NaCl + H₂O
Calculating ΔH ionization for H₃PO₃
The enthalpy of ionization of H₃PO₃ (ΔH_ionization) can be derived using Hess's Law. The total heat change for the dissociation of H₃PO₃ can be expressed as:
ΔH₃PO₃ = ΔH_ionization + ΔHCl
By rearranging the formula, we get:
ΔH_ionization = ΔH₃PO₃ - ΔHCl
Substituting the Values
- ΔH_ionization = (-106.68 kJ) - (-55.84 kJ)
- ΔH_ionization = -106.68 + 55.84
- ΔH_ionization = -50.84 kJ
Since we need the positive value for ionization, we take the absolute value:
Final Result
- ΔH_ionization = 5 kJ
Thus, the enthalpy of ionization of H₃PO₃ into its ions is approximately 5 kJ.
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Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. Can you explain this answer?
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Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. 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 Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. Can you explain this answer?.
Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. 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 Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Enthalpy of neutralization of H3PO3acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisationof H3PO3into its ions. (in KJ)Correct answer is '5'. Can you explain this answer?.
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