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Chemistry Question 74 – JEE-MAIN 2026

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Consider the reaction 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) The magnitude of enthalpy change for the reaction in kJ mol−1 is _______. (Nearest integer) Given : ΔfH∘(H2S)=−20.1 kJ mol−1 ΔfH∘(H2O)=−286.0 kJ mol−1 ΔfH∘(SO2)=−297.0 kJ mol−1

The enthalpy change for a reaction can be calculated using the standard enthalpies of formation of reactants and products.

Step 1: Identify the formula for enthalpy change✦ Active

The enthalpy change for a reaction (ΔrH∘) can be calculated using the standard enthalpies of formation (ΔfH∘) of products and reactants. The formula is:

ΔrH∘=∑npΔfH∘(products)−∑nrΔfH∘(reactants)

where np and nr are the stoichiometric coefficients of products and reactants, respectively. Note that the standard enthalpy of formation for an element in its standard state, like O2(g), is 0 kJ mol−1.

Step 2: Substitute values and calculate ΔrH∘○ Expand

For the given reaction 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g), substitute the given values:

ΔrH∘=[2×ΔfH∘(H2O)+2×ΔfH∘(SO2)]−[2×ΔfH∘(H2S)+3×ΔfH∘(O2)]
ΔrH∘=[2×(−286.0)+2×(−297.0)]−[2×(−20.1)+3×(0)]
ΔrH∘=[−572.0−594.0]−[−40.2]
ΔrH∘=−1166.0+40.2=−1125.8 kJ mol−1
💡 Teacher's Secret Hint

Pay close attention to the stoichiometric coefficients and the signs of the enthalpy values.

Step 3: Determine the magnitude and round to the nearest integer○ Expand

The question asks for the magnitude of the enthalpy change, which is the absolute value of ΔrH∘:

Magnitude=|−1125.8|=1125.8

Rounding to the nearest integer, the magnitude of the enthalpy change is 1126.

💡 Teacher's Secret Hint

Ensure to take the absolute value for 'magnitude' and then round correctly.

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