Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ - ∑n(reactants)Δ
where n = number of moles
Δ= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
Answer:
The answer to the question is;
The equilibrium constant for the reaction is 0.278
Reversibility.
Explanation:
Initial concentration = 0.500 M N₂ and 0.800 M H₂
N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)
One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia
That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂
0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂
That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂
Therefore at equilibrium we have
Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M
Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M
Number of moles of Ammonia = 0.150 M
K =
= 0.278
The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.