1. Tardigrade
  2. Question
  3. Chemistry
  4. Reaction of methanol with dioxygen was carried out and Δ H was found to be - 726 kj mol-1 at 298 K. The enthalpy change for the reaction will be CH3OH(l)+(3/2)O2(g) → CO2(g)+2H2O(l); Δ H = - 726 kj mol-1

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Q. Reaction of methanol with dioxygen was carried out and $\Delta H$ was found to be $- 726\, kj\, mol^{-1}$ at $298\, K$. The enthalpy change for the reaction will be
$CH_{3}OH_{\left(l\right)}+\frac{3}{2}O_{2\left(g\right)} \to CO_{2\left(g\right)}+2H_{2}O_{\left(l\right)}$;
$\Delta H = - 726\,kj\,mol^{-1}$

Thermodynamics

Solution:

$\Delta n_{g} = 1 - \frac{3}{2} =-\frac{1}{2}$
$\Delta H = \Delta U+\Delta n_{g}RT$
$= -726\times 1000 \,j\,mol^{-1} +\left(-\frac{1}{2}\times 8.314\,j\,K^{-1}\,mol^{-1}\times298\,K \right)$
$= -727238.78\, J \,mol^{-1} = - 727\, kj\, mol^{-1}$