Thermodynamics JEE 2019: Standard Enthalpy of Formation

Which reactions in thermodynamics JEE 2019 represent standard enthalpy of formation?

B and D are correct in this thermodynamics JEE 2019 question. From JEE Advanced 2019, Paper 1, Chemistry, the task is to identify equations whose standard reaction enthalpy directly equals the standard formation enthalpy of the product. More than one option can be correct. Judge each equation as written, including its physical states and coefficients.

A)

B)

C)

D)

Answer: B and D. A uses the wrong reference state of carbon. C forms two moles of water rather than one. Apply both checks to the original equations, not to corrected versions.

What two conditions must a standard formation reaction satisfy?

Standard enthalpy of formation is the enthalpy change when exactly one mole of a substance forms from its constituent elements in their most stable standard states. Both conditions must hold in the equation as written. This is the definition-based method used in the official solution.

1. Check the reactants: Are they the constituent elements in their stable standard states?
2. Check the product amount: Does the written equation produce exactly one mole of the specified substance?

Fractional reactant coefficients are allowed. The one-mole requirement concerns the product being formed, not each reactant. No numerical enthalpy values are needed: this question tests the definition and the equation as written.

Why is A incorrect but B correct?

A forms one mole of ethane but uses gaseous carbon instead of graphite. B forms one mole of ozone from oxygen in its stable elemental reference form, so it satisfies both conditions.

For A, check the product first. Ethane has coefficient one, so the product-amount condition passes. The failure is gaseous carbon: carbon’s stable standard state for this question is solid graphite. The repaired formation equation is:

This corrected equation qualifies; the original A does not. Correcting the carbon state changes the reaction, not just its notation.

For B, the fractional coefficient balances oxygen correctly. The atom count is:

The reactant supplies the three oxygen atoms required per ozone molecule, and the equation forms one mole of gaseous ozone. Gaseous dioxygen is oxygen’s stable elemental reference form, so:

The product need not itself be the stable reference form. Ozone has a formation enthalpy even though it is an elemental allotrope rather than a compound. B is correct.

Why is C incorrect but D correct?

C uses the correct elemental states but forms two moles of liquid water. D uses the stable standard elemental forms specified in the official solution and forms exactly one mole of sulfur dioxide. Its fractional sulfur coefficient is valid.

For C, hydrogen gas and oxygen gas pass the reference-state check. The product coefficient is two, however, so:

Divide every coefficient by two to obtain the formation equation. The reaction enthalpy must also be divided by two:

C as written is incorrect. Its enthalpy must first be rescaled to give the standard formation enthalpy of liquid water.

For D, solid molecular sulfur and gaseous dioxygen are the stable standard elemental forms used in the official solution. The sulfur balance is:

That supplies the sulfur in one mole of sulfur dioxide. One mole of oxygen molecules supplies two moles of oxygen atoms, matching the product’s oxygen content. The product coefficient is one, so D passes both checks:

Final answer: B and D.

What reasoning mistake makes option C look correct?

Checking only the elemental states wrongly admits C. Water is made from hydrogen and oxygen in their standard states, but that is only half the definition. The missing check is how many moles of water the written equation produces.

The two-mole and one-mole equations have different enthalpies. Only the one-mole equation directly represents water’s standard formation enthalpy:

Silently normalising C while leaving its enthalpy unchanged is invalid. Multiplying or dividing a thermochemical equation scales its enthalpy by the same factor. Check reference states, count product moles, and keep enthalpy scaling attached to every coefficient change.

Which three practice questions check the same method?

These original practice questions, not additional verified PYQs, check reference states, reaction scaling and the zero-enthalpy convention. Their answers are symbolic or definitional; no tabulated enthalpy data are needed.

Question 1: Does this equation represent standard formation enthalpy?

Yes. Both elemental reactants are in their stable standard states, and exactly one mole of ethane forms. Changing gaseous carbon to graphite repairs the failure in option A.

Question 2: Express the enthalpy of this reaction in terms of the standard formation enthalpy of liquid water.

Start with one-mole formation. Its reaction enthalpy equals the standard formation enthalpy of liquid water:

Multiply the equation by two. Its enthalpy doubles too:

Then reverse it to obtain the requested reaction. Reversal changes the sign, giving:

Question 3: Must ozone’s standard formation enthalpy be zero because it contains only oxygen?

No. The zero convention applies to oxygen’s stable elemental reference form:

Ozone is a different allotrope. Its formation reaction is:

Its formation enthalpy is not assigned zero merely because it contains one element. Cover the solutions and write one deciding reason beside each original option: reference state, product amount, or both conditions satisfied.

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