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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
View written solutionFree

Correct answer: B

  1. Interpret the given thermochemical equations

    The reactions are written as heat being evolved:

    C(diamond)→C(graphite)+X kJ mol−1\text{C(diamond)} \rightarrow \text{C(graphite)} + X\ \text{kJ mol}^{-1}C(diamond)→C(graphite)+X kJ mol−1 C(diamond)+O2(g)→CO2(g)+Y kJ mol−1\text{C(diamond)} + O_2(g) \rightarrow CO_2(g) + Y\ \text{kJ mol}^{-1}C(diamond)+O2​(g)→CO2​(g)+Y kJ mol−1 C(graphite)+O2(g)→CO2(g)+Z kJ mol−1\text{C(graphite)} + O_2(g) \rightarrow CO_2(g) + Z\ \text{kJ mol}^{-1}C(graphite)+O2​(g)→CO2​(g)+Z kJ mol−1

    Since heat appears on the product side, the enthalpy changes are:

    ΔH1=−X\Delta H_1 = -XΔH1​=−X ΔH2=−Y\Delta H_2 = -YΔH2​=−Y ΔH3=−Z\Delta H_3 = -ZΔH3​=−Z

  2. Apply Hess's law

    To go from diamond to CO2CO_2CO2​, we can follow two paths:

    • Direct path: C(diamond)+O2→CO2\text{C(diamond)} + O_2 \rightarrow CO_2C(diamond)+O2​→CO2​ with enthalpy change −Y-Y−Y.

    • Indirect path: first convert diamond to graphite, then burn graphite: C(diamond)→C(graphite)ΔH=−X\text{C(diamond)} \rightarrow \text{C(graphite)} \qquad \Delta H=-XC(diamond)→C(graphite)ΔH=−X C(graphite)+O2→CO2ΔH=−Z\text{C(graphite)} + O_2 \rightarrow CO_2 \qquad \Delta H=-ZC(graphite)+O2​→CO2​ΔH=−Z

      Total enthalpy change: (−X)+(−Z)=−(X+Z)(-X) + (-Z) = -(X+Z)(−X)+(−Z)=−(X+Z)

    By Hess's law, both paths must have the same enthalpy change:

    −Y=−X−Z-Y = -X - Z−Y=−X−Z

  3. Rearrange

    Y=X+ZY = X + ZY=X+Z

    Hence,

    X=Y−ZX = Y - ZX=Y−Z

  4. Check options

    • A: −X=Y+Z-X = Y + Z−X=Y+Z ❌
    • B: X=Y−ZX = Y - ZX=Y−Z ✅
    • C: X=−Y+ZX = -Y + ZX=−Y+Z ❌
    • D: X=Y+ZX = Y + ZX=Y+Z ❌

Therefore, the correct option is B.

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