Respuesta :
Answer:
The standard enthalpy of formation of this isomer of [tex]C_{8}H_{18}[/tex] is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
[tex]C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)[/tex]
[tex]\Delta H^{o}_{rxn}= -5104.1kJ/mol[/tex]
The expression for the entropy change for the reaction is as follows.
[tex]\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})][/tex]
[tex]\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol[/tex]
[tex]\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol[/tex]
[tex]\Delta H^{o}_{f}(O_{2})= 0kJ/mol[/tex]
Substitute the all values in the entropy change expression.
[tex]-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol][/tex]
[tex]-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})[/tex]
[tex]\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol[/tex]
[tex]=-220.1kJ/mol[/tex]
Therefore, The standard enthalpy of formation of this isomer of [tex]C_{8}H_{18}[/tex] is -220.1 kJ/mol.
