Reaction of methanol with dioxygen was carried out and $Δ U$ was found to be $- 726 k J m o l^{- 1}$ at $298 K$ . The enthalpy change for the reaction will be
$C H_{3} O H_{(l)} + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)}$ ; $Δ U = - 726 k J m o l^{- 1}$

- 741.5 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 727 k J m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 741. 5 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 727.2 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 727 k J m o l^{- 1}$
$Δ n_{g} = 1 - \frac{3}{2} = - \frac{1}{2}$

Now,

$Δ H = Δ U + Δ n_{g} R T$

$Δ H = - 726 \times 1000 J m o l^{- 1} + (- \frac{1}{2} \times 8.314 J K^{- 1} m o l^{- 1} \times 298 K)$

$Δ H = - 727238.78 J m o l^{- 1} \approx - 727 k J m o l^{- 1}$

Suggest Corrections

Similar questions

The reaction of cyanamide, $N H_{2} C N_{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $Δ U$ was found to be $- 742.7 k J m o l^{- 1}$ at 298 K. calculate enthalpy change for the reaction at 298 K.
$N H_{2} C N_{(g)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$