Question

Be sure to answer all parts. From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l) → SOCl2(l) + SO2(g) ΔG o rxn = −75.2 kJ SO3(g) SCl2(l) SOCl2(l) SO2(g) ΔH o f (kJ/mol) −396 −50.0 −245.6 −296.8 S o (J/mol · K) 256.7 184 − 248.1

Answer 1

Answer:

Explanation:

Hrxn = -245.6-296.8+396+50.0 = -96.4 (kJ/mol)

Since Grxn = Hrxn - T*Srxn, where T = 298.15K, we have:

Srxn = (Hrxn - Grxn)/T = (-96.4+75.2)× 1000÷298.15 = -71.1 (j/K mol)

Since: Srxn = Sf (SO2) + Sf (SOCl2)- Sf (SO3) - Sf (SCl2),

or: -71.1 = 248.1 + Sf (SOCl2) - 256.7-184,

Sf (SOCl2) = 256.7+184-71.1-248.1 = 121.5 (J/mol*k)

In order to find T such that Grxn = Hrxn - T*Srxn >= 0. since Hrxn is negative and Srxn is positive, Grxn will always be less than zero. Therefore there won't be a temperature point at which the reaction is going to be non-spontaenous