Question

The enthalpy of Sodium is 235 calories. The enthalpy Chlorine is 435 calories. The enthalpy of Sodium chloride 670 joules, what is the change in enthalpy for this reaction?

Answer 1

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]

ΔH = -2446J