SATHEE: Chemical Thermodynamics Question 361

Chemical Thermodynamics Question 361

Question: The difference between the reaction enthalpy change $(Δ_{r} H)$ and reaction internal energy change $(Δ_{r} U)$ for the reaction: $2 C_{6} H_{6} (l) + 15 O_{2} (g) \overset{}{\to} 12 C O_{2} (g) + 6 H_{2} O (l)$ at $300 K$ is $(R = 8.314, J m o l^{- 1} K^{- 1})$

Options:

A) $0 J, m o l^{- 1}$

B) $2490 J m o l^{- 1}$

C) $- 2490 J m o l^{- 1}$

D) $- 7482 J m o l^{- 1}$

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Answer:

Correct Answer: D

Solution:

$Δ H = Δ U + Δ n_{g} R T$ For the reaction $Δ n_{g} = 12 - 15 = - 3$

$Δ H - Δ U = - 3 \times 8.314 \times 300$

$= - 7482 J m o l^{- 1}$

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Chemical Thermodynamics Question 361

Question: The difference between the reaction enthalpy change (ΔrH) and reaction internal energy change (ΔrU) for the reaction: 2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) at 300K is (R=8.314,Jmol−1K−1)

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Question: The difference between the reaction enthalpy change $(Δ_{r} H)$ and reaction internal energy change $(Δ_{r} U)$ for the reaction: $2 C_{6} H_{6} (l) + 15 O_{2} (g) \overset{}{\to} 12 C O_{2} (g) + 6 H_{2} O (l)$ at $300 K$ is $(R = 8.314, J m o l^{- 1} K^{- 1})$