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Chemistry Question on Thermodynamics terms

The difference between the reaction enthalpy change (ΔrH)\left( \Delta_{r}H\right) and reaction internal energy change (ΔrU)\left( \Delta_{r}U\right) for die reaction : 2C6H6(1)+15O2(g)→12CO2(g)+6H2O(1)2C_{6}H_{6} \left(1\right) +15O_{2} \left(g\right) \to 12CO_{2} \left(g\right) + 6H_{2}O\left(1\right) at 300300 K is (R=8.31 ,mol−1K−1)\left(R=8.31 \,,mol^{-1}K^{-1}\right)

A

0 J mol−10\, J \,mol^{-1}

B

2490 J mol−12490\, J\, mol^{-1}

C

−2490 J mol−1-2490\,J\,mol^{-1}

D

−7482 J mol−1-7482\,J\,mol^{-1}

Answer

−7482 J mol−1-7482\,J\,mol^{-1}

Explanation

Solution

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_{g}RT For the reaction Δng=12−15=−3\Delta n_{g} = 12 - 15 = - 3 ΔH−ΔU=−3×8.314×300\Delta H - \Delta U= - 3 \times 8.314 \times 300 =−7482 J mol−1= - 7482\, J \,mol^{-1}