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Galina-37 [17]
3 years ago
10

Use Coulomb's Law to calculate the energy of a copper ion and an oxide ion at their equilibrium ion-pair separation distance.

Chemistry
1 answer:
Tanzania [10]3 years ago
8 0
E= (231 aj/pm) (-2)(2)/77+140 Pm

E= 231 aj (-4)/217

E= -868 aj/ 217

E= -4aj







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Answer:

ΔHr = -103,4 kcal/mol

Explanation:

<u>Using:</u>

<u>AH° (kcal/mol) </u>

<u>Metano (CH) </u>

<u>-17,9 </u>

<u>Cloro (CI) </u>

<u>tetraclorometano (CCI) </u>

<u>- 33,3 </u>

<u>Acido cloridrico (HCI) </u>

<u>-22</u>

It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:

ΔHr = (ΔH products - ΔH reactants)

For the reaction:

CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)

The balanced reaction is:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

The ΔH's of formation for these compounds are:

ΔH CH₄(g): -17,9 kcal/mol

ΔH Cl₂(g): 0 kcal/mol

ΔH CCl₄(g): -33,3 kcal/mol

ΔH HCl(g): -22 kcal/mol

The ΔHr is:

-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)

<em>ΔHr = -103,4 kcal/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
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