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3 years ago

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.

E/kJ

(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⁻ +349

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

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

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Answer: The correct answer is option(A).

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Since, upward force are equal in magnitude but opposite in direction by which they will balance out each other.so, the net force car will be due to two forces $F_r$ and $F_l$

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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