JEE Main 2025 · 29 January · evening · Single correct · +4 / −1
If C (diamond ) → C (graphite) +X kJ mol⁻¹
C (diamond) +O₂( g) → CO₂( g)+Y kJ mol ⁻¹
C (graphite) +O₂( g) → CO₂( g)+Z kJ mol⁻¹
at constant temperature. Then
- (a) −X = Y + Z
- (b) X = Y − Z
- (c) X = −Y + Z
- (d) X = Y + Z
Answer and solution
Answer (b) X = Y − Z
Solution
Chemistry ma’am · 4 steps
Subtract the graphite burning equation from the diamond burning one, and the heat terms subtract too.
- Read the heat terms. Each equation writes the heat given out as a product. So X, Y and Z are amounts of heat released, and you can add or subtract these equations like algebra, heat terms included.
- Pick the combination. The target has diamond on the left and graphite on the right. Diamond is on the left in the second equation, and graphite is on the left in the third. Take the second equation minus the third.
C(diamond) + O₂ - C(graphite) - O₂ → CO₂ - CO₂ + (Y - Z) kJ mol⁻¹
- Cancel and rearrange. O₂ and CO₂ cancel. Moving graphite to the right gives C(diamond) → C(graphite) + (Y - Z) kJ mol⁻¹.
- So. This matches the first equation term for term, so X = Y - Z. That is option (b).
The idea to keep. In Hess's law, combine the equations so that every species except your target cancels. Then do exactly the same arithmetic on the heat terms.
Answer from the NTA final answer key. Solution by PrepWiser.