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

ΔS° = 121kJ , T = 1335K

Explanation:

The reaction:

SO₃(g) + SCl₂=> SOCl₂(l) + SO₂(g)

parameters                                                                                                                    

                           SO₃(g)       SCl₂(l)       SOCl₂(l)          SO₂(g)

ΔH°f(kJ/mol)          -396           -50.0         -245.6          -296.8

S°(J/mol.k)            256.7          184               -                 248.1

ΔG°r = -75.2 kJ

a).

S° of SOCl₂

ΔH°rxn = ΔH°f[Product] - ΔH°f[reactant]

ΔH°rxn = [1 mole * ΔH°f(SOCl2(l)) + 1 mol * ΔH°f(SO2(g))] - [1 mol *ΔH°f(SO3(g)) + 1 mole ΔH°f(SCl2(l))]

ΔH°rxn = [1 mole * -245.6kJ/mol + 1 mol * -296.8kJ/mol] - [1 mol * -396kJ/mol + 1 mole * -50kJ/mol]

ΔH°rxn = -96.4 kJ

Check the added file for other comprehensive solution

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol