Question

Pentane is one of the most volatile of the hydrocarbons in gasoline. The following enthalpies of formation are given for pentane: $\Delta H_f^\circ$ for $C_5H_{12} (\ell)$: -173.5 kJ mol$^{-1}$ $\Delta H_f^\circ$ for $C_5H_{12} (g)$: -146.9 kJ mol$^{-1}$ Given the boiling point of pentane is 36.1°C what is the standard entropy change of vaporization of pentane, $\Delta S_{vap}^\circ$ in J mol$^{-1}$ K$^{-1}$? A) -8.22 B) 8.22 C) 86.0 D) -86.0 E) 2.42

          Pentane is one of the most volatile of the hydrocarbons in gasoline. The following enthalpies of formation are given for pentane:
$\Delta H_f^\circ$ for $C_5H_{12} (\ell)$: -173.5 kJ mol$^{-1}$
$\Delta H_f^\circ$ for $C_5H_{12} (g)$: -146.9 kJ mol$^{-1}$
Given the boiling point of pentane is 36.1°C what is the standard entropy change of vaporization of pentane, $\Delta S_{vap}^\circ$ in J mol$^{-1}$ K$^{-1}$?
A) -8.22
B) 8.22
C) 86.0
D) -86.0
E) 2.42

Pentane is one of the most volatile of the hydrocarbons in gasoline. The following enthalpies of formation are given for pentane:
Δ Hf^∘ for C5H12 (ℓ): -173.5 kJ mol^-1
Δ Hf^∘ for C5H12 (g): -146.9 kJ mol^-1
Given the boiling point of pentane is 36.1°C what is the standard entropy change of vaporization of pentane, Δ Svap^∘ in J mol^-1 K^-1?
A) -8.22
B) 8.22
C) 86.0
D) -86.0
E) 2.42

Added by Maria A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Banhishikha Sinha

10/05/2020

Step 1

The enthalpy change of vaporization (ΔHvap) can be calculated using the enthalpies of formation of the liquid and gaseous forms of pentane. ΔHvap = ΔHf(g) - ΔHf(l) Where: ΔHvap = standard enthalpy change of vaporization ΔHf(g) = enthalpy of formation of gaseous Show more…

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Text: Pentane is one of the most volatile of the hydrocarbons in gasoline. The following enthalpies of formation are given for pentane: ΔHf(C5H12) = -173.5 kJ/mol, ΔHf(C5H10) = 146.9 kJ/mol. Given the boiling point of pentane is 36.1°C, what is the standard entropy change of vaporization of pentane (ΔSvap)? (R = 8.314 J/(mol·K), Avogadro's number = 6.022 × 10^23 mol^-1) A) -8.22 J/(mol·K) B) 8.22 J/(mol·K) C) 86.0 J/(mol·K) D) 0.98 J/(mol·K) E) 2.42 J/(mol·K)