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Can you explain the answer of this question below:
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
The answer is c.
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Can you explain the answer of this question below:The enthalpy change ...
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Can you explain the answer of this question below:The enthalpy change ...
Enthalpy change (H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K can be calculated using the formula:
H = -n × ΔHₙₒ
Where n is the number of moles of HCl used in the reaction and ΔHₙₒ is the standard enthalpy of neutralisation.
We know that the reaction between HCl and NaOH (caustic soda) is as follows:
HCl + NaOH → NaCl + H₂O
The stoichiometry of the reaction shows that 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of NaCl and 1 mole of water.
Therefore, the number of moles of HCl used in the reaction is 1.
The standard enthalpy of neutralisation, ΔHₙₒ, is the enthalpy change when 1 mole of H⁺ ions from an acid reacts with 1 mole of OH⁻ ions from a base to form 1 mole of water under standard conditions (298 K and 1 atm pressure).
The value of ΔHₙₒ for the reaction between HCl and NaOH is -57.3 kJ/mol.
Substituting the values of n and ΔHₙₒ in the formula, we get:
H = -1 × (-57.3) = 57.3 kJ/mol
Therefore, the enthalpy change (H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K is 57.3 kJ/mol.
Answer: Option C
H = -n × ΔHₙₒ
Where n is the number of moles of HCl used in the reaction and ΔHₙₒ is the standard enthalpy of neutralisation.
We know that the reaction between HCl and NaOH (caustic soda) is as follows:
HCl + NaOH → NaCl + H₂O
The stoichiometry of the reaction shows that 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of NaCl and 1 mole of water.
Therefore, the number of moles of HCl used in the reaction is 1.
The standard enthalpy of neutralisation, ΔHₙₒ, is the enthalpy change when 1 mole of H⁺ ions from an acid reacts with 1 mole of OH⁻ ions from a base to form 1 mole of water under standard conditions (298 K and 1 atm pressure).
The value of ΔHₙₒ for the reaction between HCl and NaOH is -57.3 kJ/mol.
Substituting the values of n and ΔHₙₒ in the formula, we get:
H = -1 × (-57.3) = 57.3 kJ/mol
Therefore, the enthalpy change (H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K is 57.3 kJ/mol.
Answer: Option C
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Can you explain the answer of this question below:The enthalpy change ...
It's basically the heat of neutralisation ie, the amount of heat evolved or absorbed when 1 mole of H+ is neutralised with 1 mole of OH-... for strong acid and strong base it's -57.3 KJ
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Can you explain the answer of this question below: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 KJThe answer is c.
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