back to directory
NEET CHEMISTRYThermodynamicsHard

Question

The absolute enthalpy of neutralisation of the reaction MgO(s)+2HCl(aq)ightarrowMgCl2(aq)+H2O(l)MgO(s) + 2HCl(aq) ightarrow MgCl_{2}(aq) + H_{2}O(l) will be [CBSE PMT 2005]:

A

Less than –57.33 kJ mol⁻¹

B

–57.33 kJ mol⁻¹

C

Greater than –57.33 kJ mol⁻¹

D

57.33 kJ mol⁻¹

Step-by-Step Solution

The standard enthalpy of neutralisation for the reaction between a strong acid and a strong base in dilute aqueous solution is approximately –57.33 kJ mol⁻¹, representing the formation of one mole of water from H+(aq)H^{+}(aq) and OH(aq)OH^{-}(aq) ions [Source knowledge]. However, for the reaction involving a solid basic oxide like MgO, we apply Hess's Law using the standard enthalpies of formation (ΔfH\Delta_{f}H^{\circ}) found in the sources:

  1. Identify Enthalpy Values (at 298 K): ΔfH[Mg2+(aq)]=466.85 kJ mol1\Delta_{f}H^{\circ} [Mg^{2+}(aq)] = -466.85 \text{ kJ mol}^{-1} ΔfH[H2O(l)]=285.83 kJ mol1\Delta_{f}H^{\circ} [H_{2}O(l)] = -285.83 \text{ kJ mol}^{-1} ΔfH[MgO(s)]=601.70 kJ mol1\Delta_{f}H^{\circ} [MgO(s)] = -601.70 \text{ kJ mol}^{-1} ΔfH[H+(aq)]=0 kJ mol1\Delta_{f}H^{\circ} [H^{+}(aq)] = 0 \text{ kJ mol}^{-1} (by convention)

  2. Calculate Reaction Enthalpy (ΔrH\Delta_{r}H^{\circ}): The net ionic equation is MgO(s)+2H+(aq)Mg2+(aq)+H2O(l)MgO(s) + 2H^{+}(aq) \rightarrow Mg^{2+}(aq) + H_{2}O(l). ΔrH=[ΔfH(Mg2+,aq)+ΔfH(H2O,l)][ΔfH(MgO,s)+2×ΔfH(H+,aq)]\Delta_{r}H^{\circ} = [\Delta_{f}H^{\circ}(Mg^{2+}, aq) + \Delta_{f}H^{\circ}(H_{2}O, l)] - [\Delta_{f}H^{\circ}(MgO, s) + 2 \times \Delta_{f}H^{\circ}(H^{+}, aq)] ΔrH=[(466.85)+(285.83)][(601.70)+0]\Delta_{r}H^{\circ} = [(-466.85) + (-285.83)] - [(-601.70) + 0] ΔrH=752.68(601.70)=150.98 kJ mol1\Delta_{r}H^{\circ} = -752.68 - (-601.70) = -150.98 \text{ kJ mol}^{-1}.

Since –150.98 kJ mol⁻¹ is numerically less than –57.33 kJ mol⁻¹ (more exothermic), the correct option is A.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsabsoluteenthalpyneutralisationreactionightarrow

More Thermodynamics Questions

View all

For the gas phase reaction, $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$, which of the following conditions is correct?

A.$\Delta H = 0$ and $\Delta S < 0$
B.$\Delta H > 0$ and $\Delta S > 0$
C.$\Delta H < 0$ and $\Delta S < 0$
D.$\Delta H > 0$ and $\Delta S < 0$
EasySolve

One mole of an ideal gas at $300\text{ K}$ is expanded isothermally from an initial volume of $1\text{ litre}$ to $10\text{ litres}$. The $\Delta E$ for this process is: ($R = 2\text{ cal mol}^{-1}\text{K}^{-1}$)

A.$1381.1\text{ cal}$
B.Zero
C.$163.7\text{ cal}$
D.$9\text{ lit. atm}$
EasySolve

What is the amount of work done by an ideal gas, if the gas expands isothermally from $10^{-3} \text{ m}^3$ to $10^{-2} \text{ m}^3$ at $300 \text{ K}$ against a constant pressure of $10^5 \text{ N m}^{-2}$?

A.$+270 \text{ kJ}$
B.$-900 \text{ J}$
C.$+900 \text{ kJ}$
D.$-900 \text{ kJ}$
MediumSolve

Match List-I (Equations/Conditions) with List-II (Type of processes) and select the correct option. **List-I (Equations)** (a) $K_p > Q$ (b) $T < \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$) (c) $K_p = Q$ (d) $T > \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$) **List-II (Type of processes)** (i) Non-spontaneous (ii) Equilibrium (iii) Spontaneous and endothermic (iv) Spontaneous

A.(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
B.(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
C.(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
D.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
MediumSolve

The enthalpy of neutralization of which of the following acids and bases is nearly –13.6 Kcal [Roorkee 1999]

A.HCN and NaOH
B.HCl and KOH
C.HCl and HCN
D.HCl and NH4OH
EasySolve

The heat of combustion of carbon to $\text{CO}_2$ is $-393.5 \text{ kJ/mol}$. The heat released upon the formation of $35.2 \text{ g}$ of $\text{CO}_2$ from carbon and oxygen gas is:

A.$-315 \text{ kJ}$
B.$+315 \text{ kJ}$
C.$-630 \text{ kJ}$
D.$+630 \text{ kJ}$
MediumSolve

A system is changed from state A to state B by one path and from B to A by another path. If $E_{1}$ and $E_{2}$ are the corresponding changes in internal energy, then [Pb. PMT 2001]:

A.$E_{1} + E_{2} = -ve$
B.$E_{1} + E_{2} = +ve$
C.$E_{1} + E_{2} = 0$
D.None of the above
EasySolve

The work done to contract a gas in a cylinder is 462 joules. 128 joule energy is evolved in the process. What will be the internal energy change in the process? [MP PMT 2003]

A.+ 590 joules
B.– 334 joules
C.+ 334 joules
D.– 590 joules
EasySolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →