The link above is a hacker
Answer:
The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
![C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)](https://tex.z-dn.net/?f=C_%7B8%7DH_%7B18%7D%28g%29%2B%20%5Cfrac%7B25%7D%7B2%7DO_%7B2%7D%28g%29%20%5Crightarrow%208CO_%7B2%7D%28g%29%2B9H_%7B2%7DO%28g%29)
![\Delta H^{o}_{rxn}= -5104.1kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%20-5104.1kJ%2Fmol)
The expression for the entropy change for the reaction is as follows.
![\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})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)
![\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%3D%20-241.8kJ%2Fmol)
![\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%3D%20-393.5kJ%2Fmol)
![\Delta H^{o}_{f}(O_{2})= 0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%3D%200kJ%2Fmol)
Substitute the all values in the entropy change expression.
![-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]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)
![-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D-5324.2kJ%2Fmol%20-%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29)
![\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%20%3D-5324.2kJ%2Fmol%20%2B5104.1kJ%2Fmol)
![=-220.1kJ/mol](https://tex.z-dn.net/?f=%3D-220.1kJ%2Fmol)
Therefore, The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
Answer:
The United States customary system aka USCS or USC?