The difference between the reaction enthalpy change $(\Delta _r H)$ and reaction internal energy change $(\Delta _r U)$ for the reaction $2C_6H_{6(l)} + 15O_{2(g)} \longrightarrow 12CO_{2(g)} + 6H_2O_{(l)}$ at $300 \ K$ is $....$ $J \ mol^{-1}$ $(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

  • A
    $0$
  • B
    $2490$
  • C
    $-2490$
  • D
    $-7482$

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The heat of reaction for $C_6H_{12}O_{6(s)} + 6O_{2(g)} \to 6CO_{2(g)} + 6H_2O_{(l)}$ at constant pressure is $-651 \, kcal$ at $17 \, ^oC$. Calculate the heat of reaction at constant volume at $17 \, ^oC$ in $kcal$.

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