JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
At \(25^{\circ} C\) and 1 atm pressure, the enthalpies of combustion are as given below:
| Substance | \(H _{2}\) | \(C (\) graphite \()\) | \(C _{2} H _{6}( g )\) |
| \(\frac{\Delta_{ C } H ^{\Theta}}{ kJmol ^{-1}}\) | \(-286.0\) | \(-394.0\) | \(-1560.0\) |
- A \(+54.0\, kJ \,mol ^{-1}\)
- B \(-68.0 \,kJ \,mol ^{-1}\)
- C \(-86.0\, kJ \,mol ^{-1}\)
- D \(+97.0\, kJ \,mol ^{-1}\)
Answer & Solution
Correct Answer
(C) \(-86.0\, kJ \,mol ^{-1}\)
Step-by-step Solution
Detailed explanation
\(C _{2} H _{6}( g )+\frac{7}{2} O _{2}( g ) \rightarrow 2 CO _{2}( g )+3 H _{2} O (\ell)\) \(\Delta_{ C } H \left( C _{2} H _{6}\right)=2 \Delta_{ f } H CO _{2}( g )+3 \Delta_{ f } H \left( H _{2} O , \ell\right)\) \(-\Delta_{ f } H ^{\prime}\left( C _{2} H _{6}, g \right)\)…
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