N2(g) + 3H2(g) → 2NH3(g) ΔH=-46.1 KJ/mole which one is true about the equilibrium(Keq) constant for the given reaction.?

N₃(g) + 3H₂(g) ⇌ 2NH₃(g) …… ΔH = -46.1 KJ/mole
The forward reaction is exothermic (ΔH < 0), and thus the backward reaction is endothermic.

Keq = [NH₃]² / ([N₂] [H₂]³)
When the equilibrium position shifts to the right, the numerator increases and the denominator decreases. Hence, Keq increases.
When the equilibrium position shifts to the left, the numerator decreases and the denominator increases. Hence, Keq decreases.

High temperature favors the endothermic reaction, such that the backward reaction. Therefore, when temperature increases, the equilibrium position shifts to the left and thus Keq decreases.
(A) is true, and (B) is false.

Keq is only affected by the change of temperature. Therefore, there is no effect on Keq when pressure increases or decreases.
(B) and (C) are both false.

The answer: (A) Keq decreases with increase in temperature.

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P.S.

It is known that ln(Keq) = -(ΔH/R)(1/T) + constant
where R is a constant and ΔH is nearly constant when temperature changes.

When ΔH is positive, the expression can be rewritten as:
ln(Keq) = -(constant1/T) + constant2
When temperature increases, Keq increases.
When temperature decreases, Keq decreases.

When ΔH is negative, the expression can be rewritten as:
ln(Keq) = (constant1/T) + constant2
When temperature increases, Keq decreases.
When temperature decreases, Keq increases.

According to the expression ln(Keq) = -(ΔH/R)(1/T) + constant
Keq in neither directly proportional or inversely proportional to temperature.