JEE Main · 2024 · Shift-ImediumTHERMO-083

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always −57 kJ mol−1-57\,\text{kJ mol}^{-1}. Reason (R): Enthalpy of neutralisation is the amount of heat liberated when one mole of H+\text{H}^+ ions furnished by acid combine with one mole of OH−\text{OH}^- ions furnished by base to form one mole of water.

In the light of the above statements, choose the correct answer from the options given below:

Options
  1. a

    A true, R false

  2. b

    Both true, R not correct explanation

  3. c

    Both true, R is correct explanation

    ✓
  4. d

    A false, R true

Correct Answerc

Both true, R is correct explanation

Detailed Solution

🧠 Standard enthalpy of neutralization of strong acid with strong base = −57-57 kJ/mol because the net reaction is always \ceH++OH−−>H2O\ce{H+ + OH- -> H2O} Both strong acid and strong base are fully dissociated in solution. No ionisation energy is needed. The only process is: \ceH+(aq)+OH−(aq)−>H2O(l),ΔH=−57 kJ mol−1\ce{H+(aq) + OH-(aq) -> H2O(l)}, \quad \Delta H = -57\text{ kJ mol}^{-1}

Assertion A is true — it's always −57-57 kJ/mol for strong acid + strong base. Reason R is also true and correctly explains why: 1 mol of \ceH+\ce{H+} from the acid combines with 1 mol of \ceOH−\ce{OH-} from the base. R is the correct explanation.

Answer: (c)\boxed{\text{Answer: (c)}}

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Thermodynamics) inside The Crucible, our adaptive practice platform.

Practice in The Crucible →Browse Thermodynamics chapter →

More JEE Main Thermodynamics PYQs

2025hard

Total enthalpy change for freezing of 1 mol of water at 10°C to ice at -10°C is _____. Given: fusH = x\ kJ/mol Cp[H2O(l)] = y\ J mol-1K-1 Cp[H2O(s)] = z\ J mol-1K-1

2025easy

One mole of an ideal gas expands isothermally and reversibly from 10 dm3 to 20 dm3 at 300 K. U, q and work done in the process respectively are: Given: R = 8.3 JK-1 and mol-1 In 10 = 2.3, 2 = 0.30, 3…

2023medium

Given: (A) 2CO(g) + O2(g) 2CO2(g), H1 = -x\,kJ mol-1 (B) C(graphite) + O2(g) CO2(g), H2 = -y\,kJ mol-1 The H for the reaction C(graphite) + 12O2(g) CO(g) is:

2023medium

The enthalpy change for the adsorption process and micelle formation respectively are:

2022medium

At STP, Enthalpy of combustion of benzene (l) and acetylene (g) are -3268 and -1300\,kJ mol-1 respectively. The H for 3C2H2(g) C6H6(l) is:

View all 58 Thermodynamics PYQs →