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3 years ago

Write the equilibrium constant expressions for the following equilibria.

Chemistry

2 answers:

Nady [450]3 years ago

7 0

Answer:

1. K = [CF2Cl2] [HCl]^2 / [CCl4] [HF]^2

2. K = [NH3] [HCl] / [NH4Cl]

3. K = [H2SO4] / [SO3] [H2O]

Explanation:

The equilibrium constant for a chemical reaction is simply the ratio of the concentration of the products to that of the reactants

The equilibrium constant for the reactions are written below.

1. CCl4(g) + 2HF(g) ⇌ CF2Cl2(g) + 2HCl(g)

K = [CF2Cl2] [HCl]^2 / [CCl4] [HF]^2

2. NH4Cl(s) ⇌ NH3(g) + HCl(g)

K = [NH3] [HCl] / [NH4Cl]

3. SO3(g) + H2O(1) ⇌ H2SO4(1)

K = [H2SO4] / [SO3] [H2O]

IrinaVladis [17]3 years ago

5 0

Answer:

4. K = [CF₂Cl₂] [HCl]² / [HF]² [CCl₄]

5. K = [NH₃] [HCl]

6. K = 1 / [SO₃]

Explanation:

Equilibrium constant expressions are defined as the ratio between concentrations of products and concentrations of reactants powered to its reaction quotient (Compounds in solid or liquid phase are not taken in the expression).

For the reactions:

4. CCl₄(g) + 2HF(g) ⇌ CF₂Cl₂(g) + 2HCl(g)

Equilibrium expression, K, is:

5. NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

6. SO₃(g) + H₂O(l) ⇌ H₂SO₄(l)

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<u>Answer:</u> The empirical formula for the given compound is $C_{15}H_{24}O_1$

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The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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In 44g of carbon dioxide, 12 g of carbon is contained.

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<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 1.963 g of water, $\frac{2}{18}\times 1.963=0.218g$ of hydrogen will be contained.

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<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0092 moles.

For Carbon = $\frac{0.136}{0.0092}=14.78\approx 15$

For Hydrogen = $\frac{0.218}{0.0092}=23.69\approx 24$

For Oxygen = $\frac{0.0092}{0.0092}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is $C_{15}H_{24}O_1$

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