Answer:
The reaction is not in equilibrium and it proceed in reverse direction to reach the equilibruim condition
Explanation:
We have
H₂(g) + x₂(g) ⇆ 2HX (g)
Kc = 2.4 x 10 ⁻³ at 25°C
[tex]K_c = \frac{[HX]^2}{[H_2][X_2]}[/tex]
Now we have 3L reactor and the number of mole of species are 0.150 mole H₂, 0.150 mole X₂ and 0.600 mole HX
Therefore,
[tex][H_2] = \frac{0.150}{3} \\= 0.05M[/tex]
[tex][X_2] = \frac{0.150}{3} \\= 0.05M[/tex]
[tex][HX] = \frac{0.600}{3} \\= 0.2M[/tex]
Therefore,
[tex]Q_c = \frac{[HX]^2}{[H_2][X_2]}[/tex]
[tex]Q_c = \frac{[0.200]^2}{[0.05][0.05]}\\\\=16[/tex]
Now, we calculate the free energy change for the reaction
ΔrG = RT In Q/K
[tex]= (8.314) \times (298) \times In (\frac{16}{2.4\times 10^-^3} )\\\\=+21814.7 J/mol[/tex]
As free energy is +ve so the reaction is not in equilibruim and proceed in reverse reaction to reach tje equilibruim condition
