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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q21
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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
Consider the following cases of standard enthalpy of reaction (ΔHr∘\left(\Delta \mathrm{H}_{\mathrm{r}}^{\circ}\right.(ΔHr∘​ in kJmol−1)\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)kJmol−1) C2H6( g)+72O2( g)→2CO2( g)+3H2O(l)ΔH1∘=−1550C( graphite )+O2( g)→CO2( g)ΔH2∘=−393.5H2( g)+12O2( g)→H2O(l)ΔH3∘=−286\begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_1^{\circ}=-1550 \\ & \mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta \mathrm{H}_2^{\circ}=-393.5 \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_3^{\circ}=-286 \end{aligned}​C2​H6​( g)+27​O2​( g)→2CO2​( g)+3H2​O(l)ΔH1∘​=−1550C( graphite )+O2​( g)→CO2​( g)ΔH2∘​=−393.5H2​( g)+21​O2​( g)→H2​O(l)ΔH3∘​=−286​ The magnitude of ΔHfC2H6( g)∘\Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})}^{\circ}ΔHfC2​H6​( g)∘​ is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1 (Nearest integer).
Numerical answer
View written solutionFree

Correct answer: 95

  1. We need the standard enthalpy of formation of ethane, ΔHf∘(C2H6(g))\Delta H_f^\circ\big(\mathrm{C_2H_6(g)}\big)ΔHf∘​(C2​H6​(g)).

  2. Use the relation

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

for the combustion reaction:

C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)\mathrm{C_2H_6(g)}+\frac{7}{2}\mathrm{O_2(g)} \to 2\mathrm{CO_2(g)}+3\mathrm{H_2O(l)}C2​H6​(g)+27​O2​(g)→2CO2​(g)+3H2​O(l)

with

ΔH1∘=−1550 kJ mol−1\Delta H_1^\circ=-1550\ \text{kJ mol}^{-1}ΔH1∘​=−1550 kJ mol−1
  1. Given formation enthalpies from the other two reactions:
  • For CO2(g)\mathrm{CO_2(g)}CO2​(g):
ΔHf∘(CO2)=−393.5 kJ mol−1\Delta H_f^\circ(\mathrm{CO_2})=-393.5\ \text{kJ mol}^{-1}ΔHf∘​(CO2​)=−393.5 kJ mol−1
  • For H2O(l)\mathrm{H_2O(l)}H2​O(l):
ΔHf∘(H2O(l))=−286 kJ mol−1\Delta H_f^\circ(\mathrm{H_2O(l)})=-286\ \text{kJ mol}^{-1}ΔHf∘​(H2​O(l))=−286 kJ mol−1
  • For O2(g)\mathrm{O_2(g)}O2​(g) in standard state:
ΔHf∘(O2)=0\Delta H_f^\circ(\mathrm{O_2})=0ΔHf∘​(O2​)=0
  1. Substitute into Hess's law:
−1550=[2(−393.5)+3(−286)]−ΔHf∘(C2H6)-1550 = \left[2(-393.5)+3(-286)\right]-\Delta H_f^\circ(\mathrm{C_2H_6})−1550=[2(−393.5)+3(−286)]−ΔHf∘​(C2​H6​)
  1. Calculate the product side:
2(−393.5)=−7872(-393.5)=-7872(−393.5)=−787 3(−286)=−8583(-286)=-8583(−286)=−858 −787+(−858)=−1645-787+(-858)=-1645−787+(−858)=−1645

So,

−1550=−1645−ΔHf∘(C2H6)-1550 = -1645 - \Delta H_f^\circ(\mathrm{C_2H_6})−1550=−1645−ΔHf∘​(C2​H6​)
  1. Solve for ΔHf∘(C2H6)\Delta H_f^\circ(\mathrm{C_2H_6})ΔHf∘​(C2​H6​):
−1550+1645=−ΔHf∘(C2H6)-1550 + 1645 = -\Delta H_f^\circ(\mathrm{C_2H_6})−1550+1645=−ΔHf∘​(C2​H6​) 95=−ΔHf∘(C2H6)95 = -\Delta H_f^\circ(\mathrm{C_2H_6})95=−ΔHf∘​(C2​H6​) ΔHf∘(C2H6)=−95 kJ mol−1\Delta H_f^\circ(\mathrm{C_2H_6})=-95\ \text{kJ mol}^{-1}ΔHf∘​(C2​H6​)=−95 kJ mol−1
  1. The question asks for the magnitude of this value:
∣ΔHf∘(C2H6)∣=95 kJ mol−1\left|\Delta H_f^\circ(\mathrm{C_2H_6})\right|=95\ \text{kJ mol}^{-1}​ΔHf∘​(C2​H6​)​=95 kJ mol−1

Therefore, the required integer is 959595.

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