Calculate the internal energy of 2 moles of argon gas (assuming ideal behavior) at 298 K. Suggest two ways to increase its inter
nal energy by 10 J.
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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