Respuesta :
Explanation:
a) The balance the chemical equation for the combustion reaction.
[tex]C_{10}H_8+12O_2\rightarrow 10CO_2+4H_2O[/tex]
b) The amount of heat evolved on combustion of naphthalene = Q = -25.79 kJ
Mass of naphthalene = 0.6410 g
Moles of naphthalene = [tex]\frac{0.6410 g}{128 g/mol}=0.005 mol[/tex]
0.005 moles of naphthalene gives 25.79 kilo joules of heat. then 1 mol will give:
[tex]\Delta U^o=\frac{Q}{0.005 mol}=\frac{-25.79 kJ}{0.005 mol}=-5,158 kJ/mol=-5,158,000 J/mol[/tex]
The standard change in internal energy (ΔU°) for the combustion of 1.000 mol naphthalene is -5,158 kJ/mol.
c) [tex]\Delta H^o=\Delta U^o+n_gRT[/tex]
[tex]\Delta H^o=-5,158,000 J/mol+(-2)\times 8.314 J/mol K\times 298.15 K[/tex]
[tex]\Delta H^o=-5,162,957.64 J=-5,162.96 kJ/mol[/tex]
The standard enthalpy change (ΔH°) for the same reaction is -5,162.96 kJ/mol.
d) Standard enthalpy of formation per mole of naphthalene,
Enthlapy of formation of carbon dioxide ,[tex]\Delta H_{f,CO_2}= -393.509 kJ/mol[/tex]
Enthlapy of formation of water ,[tex]\Delta H_{f,H_2O}= -258.8kJ/mol[/tex]
[tex]\Delta H=10\times \Delta H_{f,CO_2}+4\times \Delta H_{f,H_2O} -(\Delta H_{f,naph}+12\times \Delta H_{f,O_2})[/tex]
[tex]-5,162.96 kJ/mol=10\times ( -393.509 kJ/mol) +4\times (-258.8 kJ/mol) -(H_{f,naph}-12\times 0kJ/mol)[/tex]
[tex]H_{f,naph}=10\times (-393.509 kJ/mol) +4\times (-258.8 kJ/mol) +0kJ/mol+5,162.96 kJ/mol[/tex]
[tex]H_{f,naph}=-192.67 kJ/mol[/tex]
The standard enthalpy of formation per mole of naphthalene is 192.67 kJ/mol.
