For the reaction: SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g), the equilibrium constant, Kc, is 18.0 at 1200 °C. If 1.0 mole of SO2 and 2.0 moles of NO2 are placed in a 2.0 L container, what concentration of SO3 will be present at equilibrium?

Question

Chemistry

Whitaker

27 December, 10:46

For the reaction: SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g), the equilibrium constant, Kc, is 18.0 at 1200 °C. If 1.0 mole of SO2 and 2.0 moles of NO2 are placed in a 2.0 L container, what concentration of SO3 will be present at equilibrium?

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Answers (2)

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Maximo Hatfield · Answer 1

[SO3] = 0.476 M

Explanation:

Step 1: Data given

Kc = 18.0 at 1200 °C

Number of moles SO2 = 1.0 moles

Number of moles NO2 = 2.0 moles

Volume = 2.0 L

Step 2: The balanced equation

SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g)

Step 3: Calculate molarity

Molarity = moles / volume

Molarity SO2 = 1.0 mol / 2.0 L

Molarity SO2 = 0.5 M

Molarity NO2 = 2.0 mol / 2.0 L

Molarity NO2 = 1.0 M

Step 3: Initial concentrations

[SO2] = 0.5M

[NO2] = 1.0 M

[SO3] = 0M

[NO] = 0M

Step 4: Concentration at equilibrium

[SO2] = 0.5 - X M

[NO2] = 1.0 - X M

[SO3] = XM

[NO] = XM

Step 5: Calculate concentrations

Kc = [SO3][NO] / [SO2][NO2]

Kc = 18 = x² / ((0.5 - x) (1.0-x)

x = 0.476

[SO2] = 0.5 - 0.476 M = 0.024 M

[NO2] = 1.0 - 0.476 M = 0.524 M

[SO3] = XM = 0.476 M

[NO] = XM = 0.476 M

Ireland Lara · Answer 2

[SO₃] in the equilibrium, will be 0.47 M

Explanation:

First of all we state the equilibrium reaction:

SO₂ (g) + NO₂ (g) ⇌ SO₃ (g) + NO (g)

Intially we have 1 moles of sulfur dioxide and 2 moles of nitrogen dioxide.

The stoichiometry is all 1:1.

During the reaction x amount has reactend, then in the equilibrum I would have

(1-x) moles of SO₂

(2-x) moles of NO₂

and there are formed x moles of SO₃ and x moles of NO

The thing is that, we need molar concentrations, so all the values must be divided by 2L, the vessel volume

Let's make the expression for Kc:

Kc = [SO₃]. [NO] / [SO₂]. [NO₂]

Kc = x/2. x/2 / (1-x) / 2. (2-x) / 2

Kc = x²/4 / (1-x) (2-x) / 4 → (1-x) (2-x) = 2-x-2x+x² → 2-3x+x²

18 = x²/4 / 2-3x+x²/4 → we cancel the 4 → 18 = x² / 2-3x+x²

18 (2-3x+x²) = x² → 0 = 17x² - 54x + 36

Let's solve the quadratic funtion:

17 = a; - 54 = b; 36 = c → (-b + - √ (b²-4ac)) / 2a

(-54 + - √ (-54² - 4. 17. 36)) / 2.36

x1 = 0.95

x2 = 2.22

We take, the first value so the [SO₃] in the equilibrium, will be 0.95/2L = 0.47 M