Question

If the formation of NaCl proceeds according to the following equation and change in enthalpy, what is the reaction and change in enthalpy for
the decomposition of 2 moles of NaCl?
*"Na" + "%2C1"_2- "Naci"., DeltaH= -411 "kj"*
OA
B.
C.
D.
2"Na" + 2"C" --2"Naci":Deltah= .411 ")"
"2"Na" + 2"C1" --2"Naci", DeltaH= -822 "J"
*2"Naci" --2"Na" + "C"; DeltaH= 822 "kj"*
2"Naci" --2"Na" + 2"C1", "DeltaH= 411 "kj"*

Answer 1

Answer:

• 2NaCl → 2Na + Cl₂, ΔH = 822 kJ

Explanation:

The chemical equation for the formation of NaCl is:

• Na + (1/2) Cl₂ → NaCl , ΔH = - 411 kJ

That equation means that 1 mole of NaCl is formed by the reaction of 1 mole of Na and 1/2 mole of Cl₂, with a release of energy of 411 kJ.

The decomposition of NaCl is the inverse of the formation reaction; thus, you swift products and reactants and inverse the sign of the change in enthalpy:

• NaCl → Na + 1/2 Cl₂, ΔH = 411 kJ

Since you want the decomposition of 2 moles you multiply the equation and the ΔH by 2:

• 2NaCl → 2Na + Cl₂, ΔH = 822 kJ ← answer