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Debora [2.8K]
3 years ago
10

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"*
Chemistry
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

  • <em>2NaCl → 2Na + Cl₂, ΔH = 822 kJ </em>

Explanation:

The chemical <em>equation</em> for the <em>formation of NaCl</em> 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 <em>decomposition</em> of <em>NaCl</em> is the inverse of the <em>formation</em> reaction; thus, you swift products and reactants and inverse the sign of the <em>change in enthalpy:</em>

  • 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

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