Question

The difference between the reaction enthalpy change $\left( \Delta_{r}H\right)$ and reaction internal energy change $\left( \Delta_{r}U\right)$ for die reaction :
$2C_{6}H_{6} \left(1\right) +15O_{2} \left(g\right) \to 12CO_{2} \left(g\right) + 6H_{2}O\left(1\right)$
at $300$ K is $\left(R=8.31 \,,mol^{-1}K^{-1}\right)$

• A. $0\, J \,mol^{-1}$
• B. $2490\, J\, mol^{-1}$
• C. $-2490\,J\,mol^{-1}$
• D. $-7482\,J\,mol^{-1}$

Answer 1

Answer: (D)

$\Delta H = \Delta U + \Delta n_{g}RT$
For the reaction $\Delta n_{g} = 12 - 15 = - 3$
$\Delta H - \Delta U= - 3 \times 8.314 \times 300$
$= - 7482\, J \,mol^{-1}$