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Thermodynamics question

2022 · 25 Jul · Shift 1 · Q13
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Thermodynamics question

2022 · 25 Jul · Shift 1 · Q13

JEE MainChemistryThermodynamicsNumerical+4 / −1
The enthalpy of combustion of propane, graphite and dihydrogen at 298 K298 \mathrm{~K}298 K are −2220.0 kJ mol−1,−393.5 kJ mol−1-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}−2220.0 kJ mol−1,−393.5 kJ mol−1 and −285.8 kJ mol−1-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}−285.8 kJ mol−1 respectively. The magnitude of enthalpy of formation of propane (C3H8)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)(C3​H8​) is ‾\underline{\hspace{2cm}}​kJ mol−1\mathrm{kJ} \,\mathrm{mol}^{-1}kJmol−1. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 104

  1. Write the combustion reaction of propane
C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\mathrm{C_3H_8(g) + 5O_2(g) \to 3CO_2(g) + 4H_2O(l)}C3​H8​(g)+5O2​(g)→3CO2​(g)+4H2​O(l)

Given:

  • Enthalpy of combustion of propane: ΔHc(C3H8)=−2220.0 kJ mol−1\Delta H_c(\mathrm{C_3H_8}) = -2220.0\ \mathrm{kJ\,mol^{-1}}ΔHc​(C3​H8​)=−2220.0 kJmol−1
  • Enthalpy of combustion of graphite: C(graphite)+O2→CO2,ΔH=−393.5 kJ mol−1\mathrm{C(graphite) + O_2 \to CO_2},\quad \Delta H = -393.5\ \mathrm{kJ\,mol^{-1}}C(graphite)+O2​→CO2​,ΔH=−393.5 kJmol−1 Hence, ΔHf∘(CO2)=−393.5 kJ mol−1\Delta H_f^\circ(\mathrm{CO_2}) = -393.5\ \mathrm{kJ\,mol^{-1}}ΔHf∘​(CO2​)=−393.5 kJmol−1
  • Enthalpy of combustion of dihydrogen: H2+12O2→H2O(l),ΔH=−285.8 kJ mol−1\mathrm{H_2 + \tfrac12 O_2 \to H_2O(l)},\quad \Delta H = -285.8\ \mathrm{kJ\,mol^{-1}}H2​+21​O2​→H2​O(l),ΔH=−285.8 kJmol−1 Hence, ΔHf∘(H2O(l))=−285.8 kJ mol−1\Delta H_f^\circ(\mathrm{H_2O(l)}) = -285.8\ \mathrm{kJ\,mol^{-1}}ΔHf∘​(H2​O(l))=−285.8 kJmol−1
  1. Use Hess's law

For the combustion of propane,

ΔHrxn=∑νΔHf∘(products)−∑νΔHf∘(reactants)\Delta H_{rxn} = \sum \nu \Delta H_f^\circ(\text{products}) - \sum \nu \Delta H_f^\circ(\text{reactants})ΔHrxn​=∑νΔHf∘​(products)−∑νΔHf∘​(reactants)

Since ΔHf∘(O2)=0\Delta H_f^\circ(\mathrm{O_2})=0ΔHf∘​(O2​)=0,

−2220.0=[3(−393.5)+4(−285.8)]−ΔHf∘(C3H8)-2220.0 = \left[3(-393.5) + 4(-285.8)\right] - \Delta H_f^\circ(\mathrm{C_3H_8})−2220.0=[3(−393.5)+4(−285.8)]−ΔHf∘​(C3​H8​)
  1. Calculate the product side
3(−393.5)=−1180.53(-393.5) = -1180.53(−393.5)=−1180.5 4(−285.8)=−1143.24(-285.8) = -1143.24(−285.8)=−1143.2

So,

−1180.5+(−1143.2)=−2323.7-1180.5 + (-1143.2) = -2323.7−1180.5+(−1143.2)=−2323.7

Thus,

−2220.0=−2323.7−ΔHf∘(C3H8)-2220.0 = -2323.7 - \Delta H_f^\circ(\mathrm{C_3H_8})−2220.0=−2323.7−ΔHf∘​(C3​H8​)
  1. Solve for enthalpy of formation of propane
−ΔHf∘(C3H8)=−2220.0+2323.7=103.7-\Delta H_f^\circ(\mathrm{C_3H_8}) = -2220.0 + 2323.7 = 103.7−ΔHf∘​(C3​H8​)=−2220.0+2323.7=103.7 ΔHf∘(C3H8)=−103.7 kJ mol−1\Delta H_f^\circ(\mathrm{C_3H_8}) = -103.7\ \mathrm{kJ\,mol^{-1}}ΔHf∘​(C3​H8​)=−103.7 kJmol−1

Therefore, the magnitude is

∣ΔHf∘∣=103.7≈104 kJ mol−1|\Delta H_f^\circ| = 103.7 \approx 104\ \mathrm{kJ\,mol^{-1}}∣ΔHf∘​∣=103.7≈104 kJmol−1
  1. Final answer

The required nearest integer is:

104\boxed{104}104​
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