Answer:
ΔH = -2446J
Explanation:
Based on the reaction:
2 Na(s) + Cl2(g) → 2NaCl
We can find the enthalpy of this reaction using Hess's law:
The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:
ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]
ΔH(NaCl) = 670J
ΔH(Na) = 235cal * (4.184J/1cal) = 983J
ΔHCl2 = 435cal * (4.184J/1cal) = 1820J
ΔH = 2*670J - [2*983J + 1820J]
ΔH = 1340J - [3786J]
