Question

For the reaction A(g) + 2 B(g) ↔ C(g) the initial partial pressures of gases A, B, and C are all 0.109 atm. Once equilibrium has been established, it is found that PC = 0.047 atm. What is ΔG° for this reaction (in kJ/mol) at 25°C?

Answer 1

Explanation:

As the given reaction is as follows.

             $A(g) + 2B(g) \leftrightarrow C(g)$

The given data is as follows.

Initially,    $P_{A}$ = 0.109 atm,     $P_{B}$ = 0.109 atm,

                $P_{C}$ = 0.109 atm

And, at equilibrium

        $P_{C}$ = 0.047 atm

Therefore, change in $P_{C}$ will be calculated as follows.

                    0.109 - 0.047

                 = 0.062 atm

Hence, $P_{A}$ = (0.109 + 0.062) atm = 0.171 atm

$P_{B}$ = $[0.109 + (2 \times 0.062)]$ atm

                        = 0.233 atm

Now, calculate the value of $K_{p}$ as follows.

         $K_{p} = \frac{P_{C}}{(P_{A})(P_{B})^{2}}$

                    = $\frac{0.047}{(0.171) \times (0.233)^{2}}$

                    = 5.06

Also, we known that $\Delta G^{o} = -RT lnK_{p}$

Hence, calculate the value of $\Delta G^{o}$ as follows.

        $\Delta G^{o} = -RT lnK_{p}$

                      = $-8.314 J/mol K \times 298 K \times ln 5.06$

                       = $-8.314 J/mol K \times 298 K \times 1.62$

                      = -4017.05 J/mol

or,                   = -4.017 kJ/mol              (as 1 kJ = 1000 J)

Thus, we can conclude that the value of $\Delta G^{o}$ for the given reaction is -4.017 kJ/mol.