Estimate the $H$ for the following reaction using the table of bond enthalpies in your book. $($Remember$,CO2$ has $2C=O$ bonds and $O2$ is double bonded$)($the $C=O$ bond in $CO2$ is $799k\frac{J}{m}ol)$

$C2H6+\frac{7}{2}O22CO2+$

$3H2O$

Break: $1$ C$-$C $(347)6$ H$-$C $(414)$ and $7/2O=O(498\mathrm{.}7)$

Form: $4C=O(745)6H-O(460)$