Answer:
618 J/Kmol
T > 1.36 x 10³ K
Explanation:
The balanced reaction of interest is:
SO₃ (g) + SCl₂ (l) ⇒ SOCl₂ (l) + SO₂ (g)
with the data:
ΔHºf (kJ/mol) -396 -50.0 -245.6 -296.8
Sº(J/mol·K) 256.7 184 ? -248.1
ΔGº= -75.2 kJ
We know, we can find the standard change inGibb´s free energy from the equation:
ΔGºrxn = ΔHºrxn - TΔSºrxn
So we can calculate ΔHºrxn = ∑ ΔHºf prod - ΔHºreact, and substitute into this equation to solve Sº SOCl₂.
ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ
Similarly for ΔSºrxn
ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K
= -0.689 kJ /K -+ Sº SOCl₂
Plugging the values for the expression for ΔGºrxn:
-75.2 kJ = -96.4 kJ - 298 K x ( -0.689 kJ /K + Sº SCl₂ )
-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂
-184 kJ = -298 K x Sº SCl₂
0.618 kJ/molK = Sº SCl₂
= 0.618 kJ/K x 1000 J = 618 J/Kmol
For the second part we will still be using the Gibb´s free energy change equation as above , but now we will solve for T when the reaction becomes non-spontaneous.
For the reaction to become non-spontaneous ΔGº is positive, and this happens when the term TΔSº becomes greater tha ΔHº:
ΔGºrxn = ΔHºrxn - TΔSºrxn
0 = ΔHºrxn - TΔSºrxn ⇒ TΔSºrxn = ΔHºrxn
T= ΔHºrxn / ΔSºrxn
ΔSºrxn = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol
( using the value the value just calculated from above )
T = - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K
For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.
