JEE Main · 2025 · Shift-IImediumTHERMO-144

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…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

If \ceC(diamond)−>C(graphite)+X\ce{C(diamond) -> C(graphite) + X} kJ mol−1^{-1}

\ceC(diamond)+O2(g)−>CO2(g)+Y\ce{C(diamond) + O2(g) -> CO2(g) + Y} kJ mol−1^{-1}

\ceC(graphite)+O2(g)−>CO2(g)+Z\ce{C(graphite) + O2(g) -> CO2(g) + Z} kJ mol−1^{-1}

At constant temperature. Then:

Options
  1. a

    X=Y+ZX = Y + Z

  2. b

    −X=Y+Z-X = Y + Z

  3. c

    X=−Y+ZX = -Y + Z

  4. d

    X=Y−ZX = Y - Z

    ✓
Correct Answerd

X=Y−ZX = Y - Z

Detailed Solution

🧠 Hess's Law: subtract one combustion reaction from another to get the phase transition Diamond and graphite both combust to \ceCO2\ce{CO2}. The transition diamond → graphite is found by combining these two combustion pathways.

🗺️ Route (1): \ceC(diamond)+O2−>CO2+Y kJ\ce{C(diamond) + O2 -> CO2 + Y}\,\text{kJ} means ΔH1=−Y\Delta H_1 = -Y

(2): \ceC(graphite)+O2−>CO2+Z kJ\ce{C(graphite) + O2 -> CO2 + Z}\,\text{kJ} means ΔH2=−Z\Delta H_2 = -Z

Target: \ceC(diamond)−>C(graphite)\ce{C(diamond) -> C(graphite)} =Reaction (1)−Reaction (2)⇒ΔH=−Y−(−Z)=Z−Y= \text{Reaction (1)} - \text{Reaction (2)} \Rightarrow \Delta H = -Y - (-Z) = Z - Y

Since the transition evolves heat XX: ΔHtransition=−X\Delta H_{\text{transition}} = -X −X=Z−Y⇒X=Y−Z-X = Z - Y \Rightarrow X = Y - Z

Answer: (d) X=Y−Z\boxed{\text{Answer: (d) } X = Y - Z}

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