The difference between the reaction enthalpy change ( ΔrH) and reaction internal energy change ( ΔrU) for die reaction: 2C6H6 (1) +15O2 (g) → 12CO2 (g) + 6H2O(1) at 300 K is (R=8.31 ,mol-1K-1)

Question

The difference between the reaction enthalpy change ( ΔrH) and reaction internal energy change ( ΔrU) for die reaction: 2C6H6 (1) +15O2 (g) → 12CO2 (g) + 6H2O(1) at 300 K is (R=8.31 ,mol-1K-1) (A) 0 J mol-1 (B) 2490 J mol-1 (C) -2490 J mol-1 (D) -7482 J mol-1

Tardigrade Admin · Accepted Answer

Δ H = Δ U + Δ ngRT For the reaction Δ ng = 12 - 15 = - 3 Δ H - Δ U= - 3 × 8.314 × 300 = - 7482 J mol-1

Q. The difference between the reaction enthalpy change $(Δ_{r} H)$ and reaction internal energy change $(Δ_{r} U)$ for die reaction :
$2 C_{6} H_{6} (1) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (1)$
at $300$ K is $(R = 8.31, m o l^{- 1} K^{- 1})$

$0 J m o l^{- 1}$

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

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

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

$Δ H = Δ U + Δ n_{g} RT$
For the reaction $Δ n_{g} = 12 - 15 = - 3$
$Δ H - Δ U = - 3 \times 8.314 \times 300$
$= - 7482 J m o l^{- 1}$

Q. The difference between the reaction enthalpy change (Δr​H) and reaction internal energy change (Δr​U) for die reaction : 2C6​H6​(1)+15O2​(g)→12CO2​(g)+6H2​O(1) at 300 K is (R=8.31,mol−1K−1)

0Jmol−1

2490Jmol−1

−2490Jmol−1

−7482Jmol−1