Question

Upload a file showing your work for the following questions. Sulfuric acid is produced in larger amounts by weight than any other chemical An intermediate step in the production of sulfuric acid is: 2 SO2 (g) + O2 (g) 42 SO3 (g) AG° = -141.8 kJ a) (4 pts.) What is the value of AG if the partial pressures of each gas are: SO2 = 2.0 atm, O2 = 1.0 atm, SO3 = 10.0 atm. b) (4 pts.) For the conditions given part a, state the direction that the reaction shifts to reach equilibrium.

Answer 1

(a) Given,

2SO₂(g) + O₂(g) $\Leftrightarrow$ 2SO₃(g) $\Delta$ G^o = -141.8 kJ

We know that

AG = AGº + RTinQ

Here, Q = reaction quotient

PSO AG = AG RTin So, Po,

Here,

R = 8.314 J/mol-K

T = 298.15 K (room temperature or standard temperature)

P_SO2 = 2.0 atm

P_O2 = 1.0 atm

P_SO3 = 10.0 atm

Putting all the values in the above equation, we get

(10.0atm) AG = –141.8kJ +8.314J/mol · K x 298.15Kin- (2.0atm)2(1.0atm)

AG= -141.8kJ + 2478.8191 Jln (25)

AG= -141.8kJ + 2.479kJln (25)

AG= –141.8kJ + 7.979kJ

AG= -133.8kJ

Hence, the value of

AG= -133.8kJ

(b) As the value of is negative which means the reaction is spontaneous.

G

The equilibrium constant value for this reaction is calculated using

$\Delta G^o = -RTlnK_{eq}$

$lnK_{eq}=\frac{-\Delta G^o}{ RT}$

$lnK_{eq}=\frac{-(-141.8 kJ)}{8.314 J/mol\cdot K\times 298.15 K} = \frac{(141800 J)}{8.314 J/mol\cdot K\times 298.15 K}$

$lnK_{eq}=57.2$

$K_{eq}=e^{57.2} = 6.90\times 10^{24}$

Calculated Q value from that previous part,

$Q=25$

$Q<<K_{eq}$

So, the reaction move in the forward direction to reach equilibrium.  

Hence, the reaction shifts in forward direction (from left to right).

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