First, calculate the number of moles of sodium present with the given mass,
31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium
It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.
mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
mols of H2 = 0.685 mols H2
Then, at STP, 1 mol of gas = 22.4 L.
volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
volume of H2 = 15.34 L
Answer: 15.34 L
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