Answer
A)$\Delta$$H_{C6H12}$=-155,8 kJ/mol
B)$\Delta$$H_{phenol}$=-165,4kJ/mol
Work Step by Step
A)$C_{6}$$H_{12(l)}$ + 9$O_{2(g)}$-->6$CO_{2(g)}$+ 6$ H_{2}$$O_{(l)}$
$\Delta$H$^{\circ}$=6$\times$$\Delta$$H_{H2O}$+ 6$\times$$\Delta$$H_{CO2}$ -$\Delta$$H_{C6H12}$
Therefore
$\Delta$$H_{C6H12}$=6$\times$(-285,8 - 393,5) + 3920
$\Delta$$H_{C6H12}$=-155,8 kJ/mol
B)$C_{5}$$H_{5}$$OH_{(l)}$ + 15/2$O_{2(g)}$-->6$CO_{2(g)}$+ 3$ H_{2}$$O_{(l)}$
The enthalpy of combustion is:
$\Delta$H$^{\circ}$=3$\times$$\Delta$$H_{H2O}$+ 6$\times$$\Delta$$H_{CO2}$ -$\Delta$$H_{phenol}$
Thus,
$\Delta$$H_{phenol}$=-857,4 -2361+3053
$\Delta$$H_{phenol}$=-165,4kJ/mol
