Question

2. Suppose that a system initially contains 0.45 mol of S02, 0.30 mol of O2 and 0.50 mol of SO3. Find the equilibrium composition of the unbalanced reaction as outlined below. [A_G(SO2) = -300.4 kJ/mol and A G(SO3) = -370.4 kJ/mol]. [10 points] SO2 + O2 =SO

Answer 1

The unbalanced equation is

SO, +02=SO

There are 4 O atoms on the left and 3 on the right.

Hence, we will multiply 2 on both SO₂ and SO₃ to balance the equation as follows:

2.502 +02= 2.50

Now, for a generic reaction, , the standard free energy change is calculated as follows:

aA+ B=0C

AG =CX AGC) - a x A G(A) - 6 x 4 Gº(B)

Hence, for our reaction, we can write the expression free energy change as follows:

AGⓇ = 2 X AFG(SO3) - 2 x 4G(SO2) - 1x4Gº (0) → AGⓇ = 2 x (-370.4 kJ/mol) – 2 x (-300.4 kJ/mol) - 1x0 kJ/mol AG° = -740.8 kJ/mol+600.8 kJ/mol = -140 kJ/mol = -140000 J mol-

Note: above we have used the fact that free energy of formation of O₂ is zero since it is in standard state already.

Now the standard free energy change is related to the equilibrium constant as follows:

AG' = -RT In K. -4G Ink RT → Kereta

Note that the standard condition temperature, T = 298.15 K

Hence,

-140000 /mol K =e ter = e .314 / K-1 mol-1x298,15 K = 3.375 X 1024

Now, given the following initial moles

nSO, = 0.45 mol

no, = 0.30 mol

nSO4 = 0.50 mol

Now, we can create the following ICE table to find the equilibrium concentrations.

	$2SO_2$	$+O_2$	= 2.50
Initial, mol	0.45	0.30	0.50
Change, mol	-2x	-x	+2x
Equilibrium, mol	0.45-2x	0.30-x	0.50+2x

Note that all are gaseous species and share the same volume. Hence, their moles are proportional to concentrations and we can take their concentrations in calculating the equilibrium constant expression.

Hence, we can write the expression of Kc as follows from the ICE table,

S0312 = 3.375 X 1024 (0.50 +2.c) Kc = (SO2)?[02] (0.45 – 2.0)2 x 0.30 – 2) (0.50 + 2x)2 = 3.375 x 1024 x (0.45 – 2.c)2 x (0.30 – x) 0.225 mol - 2.c)2 x 0.30

Hence, the equilibrium compositions of species can be calculated as follows:

nSO, = 0.45 mol – 2 = 0.45 mol – 2 x 0.225 mol = 0 mol

no, = 0.30 mol - I = 0.30 mol-0.225 mol = 0.075 mol

nSO4 = 0,50 mol +2z = 0.50 mol+2 x 0.225 mol = 0.95 mol

Hence, finally at equilibrium we have 0 mol of SO₂, 0.075 mol of O₂ and 0.95 mol of SO_3.