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cestrela7 [59]
3 years ago
7

Given the concentrations, calculate the equilibrium constant for this reaction: 2SO2(g) +O2(g)⇌2SO3(g) At equilibrium, the molar

concentrations for reactants and products are found to be [SO2]=0.48 M, [O2]=0.40 M, and [SO3]=1.12 M. What is the equilibrium constant (Kc) for this reaction?
Chemistry
2 answers:
Zielflug [23.3K]3 years ago
6 0

Answer:

Kc=13.61

Explanation:

A chemical equilibrium is the situation in which the ratio between the amounts of reagents and products in a chemical reaction remains constant over time.

The equilibrium constant (Kc) is useful for the study of chemical equilibrium, being a constant that indicates whether the reaction is favored to the formation of products or to the formation of reactants.

Being:

aA + bB ⇔ cC + dD

where a, b, c, d are the stoichiometric coefficients of the reaction and A, B, C, D are the symbols or formulas of the different substances involved

Then:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }

where [] is the Molar concentration of each of the substances in equilibrium.

In this case, you know: 2 SO₂(g) +O₂(g)⇌2 SO₃(g)

So: Kc=\frac{[SO_{3} ]^{2} }{[SO_{2}] ^{2} *[O_{2} ]}

Being: [SO2]=0.48 M, [O2]=0.40 M, and [SO3]=1.12 M

Kc=\frac{[1.12]^{2} }{[0.48] ^{2} *[0.40 ]}

Kc=13.61

Iteru [2.4K]3 years ago
5 0

Answer:

The equilibrium constant for this reactions is 13.61

Explanation:

Step 1: Data given

Concentrations at the equilibrium are:

[SO2]=0.48 M

[O2]=0.40 M

[SO3]=1.12 M.

Step 2: The balanced equation

2SO2(g) +O2(g) ⇌ 2SO3(g)

Step 3: Calculate Kc

Kc = [SO3]²/[O2]*[SO2]²

Kc = (1.12²) / (0.40 * 0.48²)

Kc = 13.61

The equilibrium constant for this reactions is 13.61

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Answer: The empirical formula for the given compound is C_3HF_2  and molecular formula for the given compound is C_{24}H_8F_{16}

Explanation : Given,

Mass of C = 18.24 g

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Mass of F = 16.91 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{18.24g}{12g/mole}=1.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles

Moles of Fluorine = \frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles

Step 2: 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 0.51 moles.

For Carbon = \frac{1.52}{0.51}=2.98\approx 3

For Hydrogen  = \frac{0.51}{0.51}=1

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Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 3 : 1 : 2

The empirical formula for the given compound is C_3H_1F_2=C_3HF_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

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Molar mass  = 562.0 g/mol

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n=\frac{562.0}{75}=7.49\approx 8

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}

Thus, the molecular formula for the given compound is C_{24}H_8F_{16}

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Balanced <span>equation :
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