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Thermodynamics question

2004 · Shift 0 · Q2
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Thermodynamics question

2004 · Shift 0 · Q2

JEE MainChemistryThermodynamicsMCQ+4 / −1
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole is :
  1. A
    110.5 kJ
  2. B
    -110.5 kJ
  3. C
    -676.5 kJ
  4. D
    676.5 kJ
View written solutionFree

Correct answer: B

  1. Write the given combustion reactions

    • Combustion of carbon: C(s)+O2(g)→CO2(g),ΔH=−393.5 kJ mol−1\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), \qquad \Delta H = -393.5\,\text{kJ mol}^{-1}C(s)+O2​(g)→CO2​(g),ΔH=−393.5kJ mol−1

    • Combustion of carbon monoxide: CO(g)+12O2(g)→CO2(g),ΔH=−283 kJ mol−1\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g), \qquad \Delta H = -283\,\text{kJ mol}^{-1}CO(g)+21​O2​(g)→CO2​(g),ΔH=−283kJ mol−1

  2. Required reaction

    We need the enthalpy of formation of carbon monoxide: C(s)+12O2(g)→CO(g)\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)C(s)+21​O2​(g)→CO(g)

    Let its enthalpy be ΔHf\Delta H_fΔHf​.

  3. Apply Hess's law

    If the formation of CO is followed by combustion of CO, we get combustion of carbon directly:

    [C(s)+12O2(g)→CO(g)]\bigg[\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)\bigg][C(s)+21​O2​(g)→CO(g)] [CO(g)+12O2(g)→CO2(g)]\bigg[\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g)\bigg][CO(g)+21​O2​(g)→CO2​(g)]

    C(s)+O2(g)→CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)C(s)+O2​(g)→CO2​(g)

    Therefore, ΔHf+(−283)=−393.5\Delta H_f + (-283) = -393.5ΔHf​+(−283)=−393.5

  4. Calculate ΔHf\Delta H_fΔHf​

    ΔHf=−393.5+283\Delta H_f = -393.5 + 283ΔHf​=−393.5+283 ΔHf=−110.5 kJ mol−1\Delta H_f = -110.5\,\text{kJ mol}^{-1}ΔHf​=−110.5kJ mol−1

  5. Match with options

    −110.5 kJ mol−1-110.5\,\text{kJ mol}^{-1}−110.5kJ mol−1 corresponds to Option B.

Final Answer: B. −110.5 kJ mol−1-110.5\,\text{kJ mol}^{-1}−110.5kJ mol−1

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