Question

Calcium Oxide is used to remove pollutant SO2 from smokestack gases. The ΔG° of the overall reaction is -418.6 kJ. What is PSO2 in equilibrium with air (PO2 = 0.21 atm) and solid CaO?

CaO(s)+SO2 (g)+1⁄2O2 (g)⇔CaSO4 (s)

Answer 1

From the change in free energy, we can calculate the equilibrium constant by the equation ΔG°=-RT ln(K) where K is the equilibrium constant.

-418.6x10^2 = -(8.314)(298) ln (K)
K = 2.38x10^73

K is expressed as:

K = 1 / Pso2 (Po2)^0.5 = 1/ Pso2 (0.21)^0.5
 2.38x10^73 = 1/ Pso2 (0.21)^0.5
Pso2 = 9.17x10^-74 atm

Answer 2

Gibbs free energy is defined as the combination of entropies and enthalpies into a single value.

The $\rm PSO_2$ for the given reaction is $9.17 \times 10^{-74}$ atm.

Given:

$\Delta \rm G^{\circ}&=-418.6$

What is the formula for Gibbs free energy?

The Gibbs free energy is calculated as the sum of enthalpy product's temperature and entropy of the system.

Formula = $\Delta \rm G^{\circ}&= RT\;ln\;(K)$

Where,

R = 8.314, T = 298, and K = $2.38 \times 10^{73}$

Value of K can be calculated as:

$\begin {aligned} \rm K &= \rm \dfrac{1}{PSO_2}\times {PO_2^{0.5}}\\\\\rm K &= \rm \dfrac{1}{PSO_2}(0.21)^{0.5}\\\\\rm {PSO_2} &=9.17 \times 10^{-74}\end$

Thus, the value of $\rm PSO_2$ for the given reaction is $9.17 \times 10^{-74}$ atm.

Learn more about Gibbs free energy here:

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