Solution:
The response that happens during the arrangement of CH3OH(l) can be composed as:
![\[C\left( s \right)\text{ }+\text{ }2H2O\left( g \right)\text{ }+\text{ }\left( 1/2 \right)O2\left( g \right)\text{ }\to \text{ }CH3OH\left( l \right)\text{ }\ldots \text{ }\ldots \text{ }\ldots \text{ }\ldots \text{ }\ldots \text{ }\left( I \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-db64246d95dc2a6e1bb7c7be1d0a6f89_l3.png "Rendered by QuickLaTeX.com")
The response (1) can be acquired from the given responses by following the logarithmic computations as:
![\[\begin{array}{*{35}{l}} Condition\text{ }\left( ii \right)\text{ }+\text{ }2\text{ }\times \text{ }condition\text{ }\left( iii \right)\text{ }\text{ }condition\text{ }\left( I \right) \\ ~ \\ fH\theta \text{ }\left[ CH3OH\left( l \right) \right]\text{ }=\text{ }cH\theta \text{ }+\text{ }2fH\theta \text{ }\left[ H2O\left( l \right) \right]\text{ }\text{ }rH\theta \\ ~ \\ =\text{ }\left( \text{ }393\text{ }kJ\text{ }mol1\text{ } \right)\text{ }+\text{ }2\left( \text{ }286\text{ }kJ\text{ }mol1\text{ } \right)\text{ }\text{ }\left( \text{ }726\text{ }kJ\text{ }mol1\text{ } \right) \\ ~ \\ =\text{ }\left( \text{ }393\text{ }\text{ }572\text{ }+\text{ }726 \right)\text{ }kJ\text{ }mol1 \\ ~ \\ fH\theta \text{ }\left[ CH3OH\left( l \right) \right]\text{ }=\text{ }\text{ }239\text{ }kJ\text{ }mol1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bff4cd69556c9348f32d26c9deaf29d9_l3.png "Rendered by QuickLaTeX.com")
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
Calculate the standard enthalpy of formation of CH3OH(l) from the following data: