A system being studied in a 3.0-liter reactor was charged with 0.150 moles of H2, 0.150 moles of X2, and 0.600 moles of HX. A catalyst was introduced using a remote unit. Is the reaction at equilibrium

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

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,

Therefore,

Now, we calculate the free energy change for the reaction

ΔrG = RT In Q/K

As free energy is +ve so the reaction is not in equilibruim and proceed in reverse reaction to reach tje equilibruim condition

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