HI has a normal boiling point of-35.4 degree C, and its delta H_vap is 21.16 kJ/

Question

HI has a normal boiling point of-35.4 degree C, and its delta H_vap is 21.16 kJ/mol. Calculate the molar entropy of vaporization (delta S_vap). Predict the signs (positive or negative) of delta H_rxn and delta S_rxn for: H_2O (l) rightarrow H_2O (g) delataH_rxn ___________ delta S_rxn__________ Octane (C_8H_18) undergoes combustion according to the following thermochemical equation: 2C_8H_18(l) + 25 O_2(g) rightarrow 16 CO_2(g) + 18H_2O(l) delata h degree _rxn = -11, 020kJ Given that delta H degree f[CO_2(g)] = -393.5 kJ/mol and delata H degree f[H_2O(l)] = -25.8 kJ/mol, calculate the standard enthalpy of formation of octane.

Explanation / Answer

Q8.

Tb = -35.4 °C = -35.4 °C + 273° = 237.6 K

so..

dS = dH/T = (21.16*10^3)/(237.6)

dS = 89.0572J/molK

Q9

predict signs

H2O(l) --> H2O(g)

this is endothermic, the system requires energy, so dH must be possitive

since the sysmtem gets more chaotic, the entropy increases

Q10

HRxn = Hproduct s- Hreactatnts

-11020 = 16*H-CO2 + 18*H-H2O .- (2*H-opctane + 25*HO2 )

-11020 = 16*-393.5+ 18*-285.8 - (2*H-octane + 25*0)

H-octane =-( -11020 + 393.5*16 +285.8 *18 )/2

H-poctane = -210.2kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.