For the reaction N2(g) + 3H2(g) <=> 2NH3(g) with ΔH = -92 kJ/mol, state the equilibrium law and the effect of changes in concentration, temperature, and pressure on the equilibrium position.

The forward reaction is exothermic ($\Delta H = -92 \text{ kJ/mol}$).

An increase in temperature will shift the equilibrium in the endothermic direction to absorb the added heat. The reverse reaction (decomposition of $\text{NH}_3$) is endothermic.
The equilibrium position will shift to the left, favoring the formation of reactants ($\text{N}_2$ and $\text{H}_2$).

(B) Decrease in pressure:

Reactant side: $1 \text{ mol N}_2 + 3 \text{ mol H}_2 = 4 \text{ moles of gas}$
Product side: $2 \text{ mol NH}_3 = 2 \text{ moles of gas}$

A decrease in pressure will shift the equilibrium to the side with a greater number of moles of gas to counteract the pressure change.
The equilibrium position will shift to the left, favoring the formation of reactants ($\text{N}_2$ and $\text{H}_2$).

Molar mass of $\text{CH}_3\text{OH} = (1 \times 12) + (4 \times 1) + (1 \times 16) = 12 + 4 + 16 = 32 \text{ g/mol}$
Molar mass of $\text{CH}_3\text{CH}_2\text{OH} = (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 \text{ g/mol}$

Moles of methanol ($n_{\text{methanol}}$):

Moles of ethanol ($n_{\text{ethanol}}$):

Mole fraction of methanol ($X_{\text{methanol}}$):

Mole fraction of ethanol ($X_{\text{ethanol}}$):

Partial vapor pressure of methanol ($P_{\text{methanol}}$):

Partial vapor pressure of ethanol ($P_{\text{ethanol}}$):

2. (a) (i) Define the term standard enthalpy of formation.
The *standard enthalpy