Home
Class 12
CHEMISTRY
A: Enthalpy of neutralisation of 1 equiv...

A: Enthalpy of neutralisation of 1 equivalent each of `HCl ` and `H_(2)SO_(4)` with NaOH is same.
R : Enthalpy of neutralisation is always the heat evolved when 1 mole acid is neutralised by a base.

A

If both Assertion & Reason are true and the reason is the correct explanation of the assertion , then mark (1)

B

If both Assertion & Reason are true and the reason is not the correct explanation of the assertion, then mark (2)

C

If Assertion is true statement but Reason is false, then mark (3)

D

If both Assertion and Reason is false, then mark (3)

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both the assertion (A) and the reason (R) provided. ### Step 1: Understanding the Assertion (A) The assertion states that the enthalpy of neutralization of 1 equivalent each of HCl and H₂SO₄ with NaOH is the same. - **Neutralization Reaction**: - For HCl: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] - For H₂SO₄ (considering 1 equivalent): \[ \frac{1}{2} \text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \] - **Equivalents**: - 1 mole of HCl provides 1 mole of H⁺ ions. - 0.5 moles of H₂SO₄ provide 1 mole of H⁺ ions (since it can donate 2 H⁺ ions per mole). Thus, when we take 1 equivalent of H₂SO₄, we are using 0.5 moles of it, which will also provide 1 mole of H⁺ ions, similar to HCl. ### Step 2: Conclusion for Assertion (A) Since both HCl and 0.5 moles of H₂SO₄ provide the same amount of H⁺ ions for the reaction with NaOH, the enthalpy of neutralization for both cases will be the same. Therefore, assertion (A) is **True**. ### Step 3: Understanding the Reason (R) The reason states that the enthalpy of neutralization is always the heat evolved when 1 mole of acid is neutralized by a base. - **Analysis of Reason (R)**: - This statement is misleading because the enthalpy of neutralization is defined for the reaction of 1 equivalent of acid with a base, not strictly 1 mole. - In the case of diprotic acids like H₂SO₄, we can use half a mole to achieve 1 equivalent, which means the statement does not hold true in all scenarios. ### Step 4: Conclusion for Reason (R) The reason (R) is **False** because it does not account for the fact that different acids can provide different amounts of H⁺ ions, and thus the enthalpy of neutralization can be calculated based on equivalents rather than strictly moles. ### Final Conclusion - Assertion (A) is **True**. - Reason (R) is **False**.
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (Section - C) Previous Years Questions|60 Videos
  • THE SOLID STATE

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION - D) (ASSERTION-REASON TYPE QUESTION)|20 Videos

Similar Questions

Explore conceptually related problems

If enthalpy of neutralisation of HCl by NaOH is -58.84 kJ/mol and by NH_4OH is -52.26 kJ/mol, then enthalpy of ionisation of NH_4OH is

What would be the heat released when 0.5 mole of HCl in solution is neutralised by 0.25 mole of KOH?

The enthalpy of neutralisation of HCl by NaOH IS -55.9 kJ and that of HCN by NaOH is -12.1 kJ mol^(-1) . The enthalpy of ionisation of HCN is

Enthalpy of neutralisation of all strong acids and strong bases has the same value because

The enthalpy of neutralisation of a weak acid in 1 M solution with a strong base is -5.6 kJ mol^(-1) . Enthalpy of ionization of the acid is 1.5 kJ mol^(-1) and enthalpy of neutralization of the strong acid with a strong base is -57.3 kJ"equiv"^(-1) , what is the % ionization of the weak acid in molar solution (assume the acid to be monobasic) ?

Statement-1 : Neutralisation reaction is an Endothermic Process. Statement-2 : Standard Enthalpy of Neutralisation for different pairs of strong acid and strong base are different. Statement-3 : Standard Enthalpy of Neutralisation for a pair of strong acid and strong base is higher than that of weak acid and weak base.

Enthalpy of neutralisation of acetic acid by NaOh is -50.6 kJ mol^(-1) . Calculate DeltaH for ionisation of CH_(3)COOH . Given. The heat of neutralisation of a strong acid with a strong base is -55.9 kJ mol^(-1) .

1 mole of H_(2)SO_(4) will exactly neutralise:

Enthalpy of neutralisation of acetic acid by NaOH is -50.6KJ//mol and the heat of neutralisation of a storng acid with a strong bases is -55..9KJ//mol . What is the value of DeltaH for the ionisation of CH_(3)COOH ?

The heat of neutralisation for strong acid and strong base forming 2 moles of water is