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The enthalpy change (∆H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K is
  • a)
    -68 KJ
  • b)
    -65 KJ
  • c)
    -057.1 KJ
  • d)
    -50 KJ
Correct answer is option 'C'. Can you explain this answer?
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The enthalpy change (∆H) for neutralisation of 1 M HCl by causti...
It's actually factual.The enthalpy of neutralisation for strong acid and strong base per mole is 57.1KJ i.e 0.0571J/mole
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The enthalpy change (∆H) for neutralisation of 1 M HCl by causti...
It's actually factual.The enthalpy of neutralisation for strong acid and strong base per mole is 57.1KJ i.e 0.0571J/mole
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The enthalpy change (∆H) for neutralisation of 1 M HCl by causti...
Enthalpy Change for Neutralization of HCl by Caustic Soda

Introduction:
The enthalpy change (H) for the neutralization of HCl by caustic soda (NaOH) can be determined using the equation:

HCl + NaOH → NaCl + H2O

This reaction is an example of an acid-base neutralization reaction, where an acid (HCl) reacts with a base (NaOH) to form a salt (NaCl) and water (H2O).

Enthalpy Change:
The enthalpy change for this reaction can be determined using the concept of enthalpy of neutralization, which is defined as the enthalpy change that occurs when one mole of an acid reacts with one mole of a base to form one mole of water under standard conditions (298 K and 1 atm).

The enthalpy change for neutralization of HCl by caustic soda can be represented as:
H = ΔH/ν

where:
ΔH = enthalpy change in kJ
ν = stoichiometric coefficient of the acid/base in the balanced equation

In the balanced equation, the stoichiometric coefficient of HCl is 1, and the stoichiometric coefficient of NaOH is also 1. Therefore, ν = 1 for both the acid and the base.

Calculation:
Given that the concentration of HCl is 1 M, it means that 1 mole of HCl reacts with 1 mole of NaOH.

Since the enthalpy change is given as -57.1 kJ, the enthalpy change for neutralization of 1 mole of HCl by 1 mole of NaOH is -57.1 kJ.

Therefore, the correct answer is option 'C' (-57.1 kJ).

Conclusion:
The enthalpy change for the neutralization of 1 M HCl by caustic soda in dilute solutions at 298 K is -57.1 kJ. This value represents the heat released during the reaction and indicates that the reaction is exothermic.
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The enthalpy change (∆H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K isa)-68 KJb)-65 KJc)-057.1 KJd)-50 KJCorrect answer is option 'C'. Can you explain this answer?
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Solutions for The enthalpy change (∆H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K isa)-68 KJb)-65 KJc)-057.1 KJd)-50 KJCorrect 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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