Question

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 S o of SOCl2(l) = J/mol · K T = × 10 K Enter your answer in scientific notation.

Answer 1

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.