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Thermodynamics question

2025 · 29 Jan · Shift 2 · Q15
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  5. /2025 · 29 Jan · Shift 2 · Q15

Thermodynamics question

2025 · 29 Jan · Shift 2 · Q15

JEE MainChemistryThermodynamicsMCQ+4 / −1
If C\quad CC(diamond )→C) \rightarrow C)→C(graphite) +X kJ mol−1+X \mathrm{~kJ} \mathrm{~mol}^{-1}+X kJ mol−1 C (diamond) +O2( g)→CO2( g)+YkJmol−1+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Y} \mathrm{kJ} \mathrm{mol}{ }^{-1}+O2​( g)→CO2​( g)+YkJmol−1 C (graphite) +O2( g)→CO2( g)+ZkJmol−1+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}+O2​( g)→CO2​( g)+ZkJmol−1 at constant temperature. Then
  1. A
    −X = Y + Z
  2. B
    X = Y − Z
  3. C
    X = −Y + Z
  4. D
    X = Y + Z
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