For the reaction C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 CO 2 ( g ) + 4 H 2 O ( l ) at constant temperature, Δ H − Δ E is:NEET 2003

C 3 H 8 ( g ) + 5 0 2 ( g ) Δ n Δ H → 3 CO 2 ( g ) + 4 H 2 O ( i ) = 3 − 6 = − 3 = Δ E + P Δ V or Δ H − Δ E = P Δ V Δ H − Δ E = Δ nRT = − 3 RT

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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionMedium2 minPYQ_2003

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For the reaction $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (l)$ at constant temperature, $Δ H - Δ E$ is:NEET 2003

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Answer:

Solution:

$C_{3} H_{8 (g)} + 5 0_{2 (g)} Δ n Δ H \to 3 CO_{2 (g)} + 4 H_{2} O_{(i)} = 3 - 6 = - 3 = Δ E + P Δ V$ or $Δ H - Δ E = P Δ V$ $Δ H - Δ E = Δ nRT = - 3 RT$

Stream:NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2003

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