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elena55 [62]
3 years ago
13

Given the following thermodynamic data, calculate the lattice energy of LiCl:

Chemistry
1 answer:
tiny-mole [99]3 years ago
3 0

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

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Please bear in mind that wurtz reaction fails whenever tertiary alkyl halides are used.

An example of Wurtz reaction is given below:

2R – X + 2Na → R–R + 2Na + X−

<h3>What are organic compounds?</h3>

Organic compounds can simply be defined as those classes of organic molecules which contain carbon atoms covalently bonded to hydrogen atoms (C-H bonds).

Below are some few general characteristics of organic compounds:

  • All organic compounds contain carbon.

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Some classes of organic compounds are:

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So therefore, Wurtz reaction is a special type of organic reaction involving the synthesis of aliphatic hydrocarbons from two molecules of an alkyl halide and two atoms of sodium in the presence of dry ether solution

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5 0
2 years ago
If nitrogen and hydrogen combine in a combustion reaction, what would the product of the reaction be?
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4 0
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List the spectator ions in the following reaction.
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Answer:

2Na⁺ (aq) and 2OH⁻(aq)

Explanation:

Spectator ions:

Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.

Given ionic equation:

Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)

In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq)  net ionic equation can be written as,

Net ionic equation:

Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)

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3 years ago
A student wants to make a 0.600 M aqueous solution of barium sulfate, BaSO4, and has a bottle containing 12.00 g of barium sulfa
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<u>Answer: </u>The volume of the solution is 85.7 mL

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}} .....(1)

We are given:

Molarity of solution = 0.600 M

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Putting values in equation 1, we get:

0.600=\frac{12.00\times 1000}{233.4\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{12.00\times 1000}{233.4\times 0.600}=85.68mL=85.7mL

The rule of significant number that is applied for the problems having multiplication and division:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

Here, the least number of significant figures is 3 that is determined by the number, 0.600. Thus, the answer must have these many significant figures only.

Hence, the volume of the solution is 85.7 mL

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7) An ionic bond could form between:
laila [671]

Answer:

the answer is nitrogen and rubidium (C)

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