Question

(eao 20) anoiteu For 52 FDIDD:C CHo1ce Oneaou bopra eg 2) 2unot old:Toibolno 2. (9 pts) For the reaction under 25°C and 1.0 atm, AHo is -125 kJ. 2 C4H10 (g) 13 O2 (g) 8 CO2 (g) 10 H20 (g) (a) What is the AS° in J/K? Substance C4H10 (g) 02 (g) QURTRV So (J/K mol) 310.0 205.0 213.6 CO2 (g) 188.83 H20 (g) (b) Calculate AG°. Is the reaction spontaneous under these conditions? (c) What is the equilibrium constant, K at these conditions? (R 8.314 J/mol K)

Answer 1

Q2. (a) $\Delta$ S^o = S^o products - S^o reactants

$\Delta$S^o = [8 * S^o CO₂ (g) + 10 * S^o H₂O (g)] - [2 * S^o C₄H₁₀ (g) + 13 * S^o O₂ (g)]

$\Delta$S^o = [8 * (213.6 J/K.mol) + 10 * (188.83 J/K.mol)] - [2 * (310.0 J/K.mol) + 13 * (205.0 J/K.mol)]

$\Delta$S^o = 1708.8 J/K + 1888.3 J/K - 620.0 J/K - 2665 J/K

$\Delta$S^o = 312.1 J/K

(b) $\Delta$ G^o = $\Delta$ H^o - (T) * ($\Delta$S^o)

where T = standard temperature = 298 K

$\Delta$G^o = (-125 kJ) - (298 K) * (312.1 J/K) * (1 kJ / 1000 J)

$\Delta$G^o = -125 kJ - 93 kJ

$\Delta$G^o = -218 kJ

Since $\Delta$ G^o is less than zero, therefore reaction is spontaneous.

(c) ln(K) = -($\Delta$G^o) / (R * T)

where R = constant = 8.314 J/mol-K

T = standard temperature = 298 K

K = equilibrium constant

ln(K) = -(-218 x 10³ J) / [(8.314 J/mol-K) * (298 K)]

kn(K) = 88

K = e⁸⁸

K = 1.64 x 10³⁸