Calculate `Delta_(r )H` for the following homogeous gaseous reaction :
`CH_(3)COCH_(3)(g)+2O_(2)(g)rarr CH_(3)COOH(g)+CO_(2)(g)+H_(2)O(g)`
Use the data in kcal `mol^(-1)`.
`Delta_("bond C-H")=99 , Delta_("bond C - C")=83 , Delta_("bond C = O")=173` ,
`Detla+("bond O = O")=118 , Delta_("bond C - O")=84 , Delta_("bond O - H")=110`

Explanation

To calculate the standard enthalpy change (Δ_rH) for the reaction:

\[ \text{CH}_3\text{COCH}_3(g) + 2\text{O}_2(g) \rightarrow \text{CH}_3\text{COOH}(g) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \]

we will use the bond energies provided. The formula to calculate Δ_rH using bond energies is:

\[ \Delta_rH = \text{(Total bond energy of reactants)} - \text{(Total bond energy of products)} \]

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