Enthalpy of Neutralisation NEET 2005: MgO Reaction Explained

Q: What is the answer to the enthalpy of neutralisation NEET 2005 question?

The official answer is option A: less than −57.33 kJ/mol. The supplied formation enthalpies give approximately −151 kJ/mol for MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l). Since −151 is more negative than −57.33, the signed reaction enthalpy is lower.

Q: Why is the enthalpy of the MgO and HCl reaction not −57.33 kJ/mol?

The usual −57.33 kJ/mol value describes dilute aqueous H⁺ and OH⁻ combining to form one mole of water. MgO is a solid basic oxide, so its reaction also includes conversion of solid MgO into aqueous magnesium ions. The standard aqueous neutralisation shortcut therefore does not apply.

Q: How do you calculate the enthalpy of the MgO and HCl reaction?

Write the net ionic equation as MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l). Using the supplied formation enthalpies, products minus reactants gives [−467 + (−286)] − [−602 + 2(0)] = −151 kJ/mol. The zero for aqueous H⁺ follows the conventional aqueous-ion reference.

Q: Should I multiply −57.33 by 2 because MgO reacts with 2HCl?

No: the reaction consumes two hydrogen ions but forms only one mole of water per mole of MgO. It is also not the standard aqueous H⁺–OH⁻ neutralisation reaction. Use the full formation-enthalpy calculation rather than multiplying the standard neutralisation value by the HCl coefficient.

What is the answer to the enthalpy of neutralisation NEET 2005 question?

The official answer is A for the enthalpy of neutralisation NEET 2005 question. The supplied calculation gives this approximate enthalpy change for the reaction as written:

Δ H \approx - 151 k J m o l^{- 1}

The task is to compare the following reaction’s enthalpy with the usual strong acid–strong base neutralisation value. The deciding observation is solid magnesium oxide, not an aqueous base already supplying hydroxide ions.

M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)

The supplied choices are:

A: Less than

- 57.33 k J m o l^{- 1}

B: Equal to

- 57.33 k J m o l^{- 1}

C: Greater than

- 57.33 k J m o l^{- 1}

+ 57.33 k J m o l^{- 1}

This belongs to Chemical Thermodynamics. The question bank tags it hard and sets a 60-second target; neither label measures actual student performance.

Why must we write the net ionic reaction first?

The standard shortcut applies to aqueous hydrogen and hydroxide ions forming water. Here, a solid basic oxide reacts with acid, so the chemical change is different. Establish the net ionic equation before assigning an enthalpy value.

Step 1: Identify what the standard value describes.

For ordinary dilute aqueous strong acid–strong base neutralisation, the net reaction and enthalpy are:

H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l)

Δ H \approx - 57.33 k J m o l^{- 1} of water formed

Step 2: Check the physical state of the base.

Magnesium oxide is a solid basic oxide, not an aqueous strong base supplying pre-existing hydroxide ions. Split the aqueous strong electrolytes into ions, but retain the solid:

M g O (s) + 2 H^{+} (a q) + 2 C l^{-} (a q) \to M g^{2 +} (a q) + 2 C l^{-} (a q) + H_{2} O (l)

Step 3: Cancel spectator chloride ions.

M g O (s) + 2 H^{+} (a q) \to M g^{2 +} (a q) + H_{2} O (l)

This reaction forms hydrated magnesium ions as well as water. Hydration alone does not establish the final enthalpy; the formation-enthalpy calculation accounts for the full reaction. Two hydrogen ions are consumed, but only one water molecule is formed.

How does the formation-enthalpy calculation give option A?

The supplied formation enthalpies give a reaction enthalpy more negative than the standard neutralisation value, selecting A. Keep two tasks separate: calculate products minus reactants, then compare the signed numbers.

Step 4: Write the formation-enthalpy relation.

Δ H_{r e a c t i o n}^{\circ} = \sum_{p r o d u c t s} ν Δ H_{f}^{\circ} - \sum_{r e a c t a n t s} ν Δ H_{f}^{\circ}

The coefficients multiply the formation enthalpies. Use these supplied approximate values:

Aqueous magnesium ions:

Δ H_{f}^{\circ} [M g^{2 +} (a q)] \approx - 467 k J m o l^{- 1}

Liquid water:

Δ H_{f}^{\circ} [H_{2} O (l)] \approx - 286 k J m o l^{- 1}

Solid magnesium oxide:

Δ H_{f}^{\circ} [M g O (s)] \approx - 602 k J m o l^{- 1}

The standard formation enthalpy of aqueous hydrogen ions is zero by the conventional aqueous-ion reference. This is a reference choice, not a claim that hydrogen ions cannot participate in energy changes.

Δ H_{f}^{\circ} [H^{+} (a q)] = 0

Step 5: Substitute, including the hydrogen-ion coefficient.

The two hydrogen ions contribute twice zero:

Δ H^{\circ} \approx [- 467 + (- 286)] - [- 602 + 2 (0)]

Δ H^{\circ} \approx - 753 - (- 602) = - 753 + 602 = - 151 k J m o l^{- 1}

Step 6: Compare signed enthalpies. $- 151 < - 57.33$

The reaction is more exothermic, so its signed enthalpy is less than the reference value: option A. This approximate result is per mole of reaction as written, which also forms one mole of water.

The source uses “absolute”, but its options and official solution compare signed enthalpy. Do not use the positive magnitude below to select the option:

| Δ H | \approx 151 k J m o l^{- 1}

Why is option B wrong even though magnesium oxide is basic?

Option B assigns the correct standard value to the wrong net reaction. The incorrect method is to identify hydrochloric acid as strong, identify magnesium oxide as basic, and immediately choose the standard neutralisation value. The missing check is whether aqueous hydrogen and hydroxide ions are the actual reacting species.

Compare the two net ionic equations:

H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l)

M g O (s) + 2 H^{+} (a q) \to M g^{2 +} (a q) + H_{2} O (l)

The shortcut leaves out the conversion of solid magnesium oxide into aqueous magnesium ions. The enthalpy calculation must include that change.

The coefficient of two for hydrochloric acid does not justify this calculation:

2 (- 57.33) = - 114.66 k J

Only one mole of water forms, and this is not standard aqueous-ion neutralisation. Use this checking sequence:

Inspect physical states.
Write the net ionic equation.
Count the water formed.
Choose the thermochemical method.

The MgO result follows from the supplied values. It does not establish that every solid basic oxide must give a more negative enthalpy.

How do you solve three related Chemical Thermodynamics practice questions?

Use the standard value only after confirming the aqueous hydrogen–hydroxide reaction. Scale total heat with the amount of water formed, and reverse the enthalpy sign when reversing a reaction. These are original related practice questions, not verified past-paper questions.

What is the heat change when dilute aqueous HCl and NaOH form one mole of water?

Related practice question 1: The enthalpy change is approximately: $Δ H \approx - 57.33 k J$

Separate the aqueous electrolytes into ions. Sodium and chloride ions appear unchanged on both sides:

H^{+} (a q) + C l^{-} (a q) + N a^{+} (a q) + O H^{-} (a q) \to N a^{+} (a q) + C l^{-} (a q) + H_{2} O (l)

Cancel those spectator ions:

H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l)

This matches the standard reaction. Since one mole of water forms, the total enthalpy change has the same numerical value as the molar neutralisation enthalpy.

What heat change accompanies formation of two moles of water?

Related practice question 2: Under the same dilute strong acid–strong base conditions, the total heat change doubles. Multiply the molar value by the amount of water formed:

Δ H = 2 m o l \times (- 57.33 k J m o l^{- 1}) = - 114.66 k J

The molar neutralisation enthalpy does not double. It remains:

Δ H_{n e u t} \approx - 57.33 k J m o l^{- 1} of water

What is the enthalpy change when the magnesium oxide reaction is reversed?

Related practice question 3: The reverse reaction has a positive enthalpy change. Reversing a reaction reverses its enthalpy sign without changing its magnitude:

M g^{2 +} (a q) + H_{2} O (l) \to M g O (s) + 2 H^{+} (a q)

Δ H_{r e v e r s e} = - Δ H_{f o r w a r d} \approx - (- 151) = + 151 k J m o l^{- 1}

This approximate result is per mole of the reverse reaction as written. It absorbs the energy released by the forward process.

Identify the reaction first; apply the value and sign convention second.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see Series LCR Circuit NEET 2026: Peak Current and Resonance.

Frequently asked questions

What is the answer to the enthalpy of neutralisation NEET 2005 question?

The official answer is option A: less than −57.33 kJ/mol. The supplied formation enthalpies give approximately −151 kJ/mol for MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l). Since −151 is more negative than −57.33, the signed reaction enthalpy is lower.

Why is the enthalpy of the MgO and HCl reaction not −57.33 kJ/mol?

The usual −57.33 kJ/mol value describes dilute aqueous H⁺ and OH⁻ combining to form one mole of water. MgO is a solid basic oxide, so its reaction also includes conversion of solid MgO into aqueous magnesium ions. The standard aqueous neutralisation shortcut therefore does not apply.

How do you calculate the enthalpy of the MgO and HCl reaction?

Write the net ionic equation as MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l). Using the supplied formation enthalpies, products minus reactants gives [−467 + (−286)] − [−602 + 2(0)] = −151 kJ/mol. The zero for aqueous H⁺ follows the conventional aqueous-ion reference.

Should I multiply −57.33 by 2 because MgO reacts with 2HCl?

No: the reaction consumes two hydrogen ions but forms only one mole of water per mole of MgO. It is also not the standard aqueous H⁺–OH⁻ neutralisation reaction. Use the full formation-enthalpy calculation rather than multiplying the standard neutralisation value by the HCl coefficient.

Enthalpy of Neutralisation NEET 2005: MgO and HCl

What is the answer to the enthalpy of neutralisation NEET 2005 question?

Why must we write the net ionic reaction first?

How does the formation-enthalpy calculation give option A?

Why is option B wrong even though magnesium oxide is basic?

How do you solve three related Chemical Thermodynamics practice questions?

What is the heat change when dilute aqueous HCl and NaOH form one mole of water?

What heat change accompanies formation of two moles of water?

What is the enthalpy change when the magnesium oxide reaction is reversed?

Frequently asked questions

What is the answer to the enthalpy of neutralisation NEET 2005 question?

Why is the enthalpy of the MgO and HCl reaction not −57.33 kJ/mol?

How do you calculate the enthalpy of the MgO and HCl reaction?

Should I multiply −57.33 by 2 because MgO reacts with 2HCl?

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