106k views
5 votes
For the chemical equation

SO₂(g) + NO₂(g) SO₂(g) + NO(g)
the equilibrium constant at a certain temperature is 2.60. At this temperature, calculate the number of moles of NO₂(g) that
must be added to 2.60mol SO₂ (g) in order to form 1.20 mol SO, (g) at equilibrium.

User Zava
by
7.2k points

1 Answer

4 votes

The balanced chemical equation for the given reaction is:

SO₂(g) + NO₂(g) ⇌ SO₂(g) + NO(g)

The equilibrium constant for the reaction at a certain temperature is given as Kc = 2.60.

Now, let's assume that x moles of NO₂(g) react with 2.60 mol of SO₂(g) to form 1.20 mol of SO(g) at equilibrium. Then, the equilibrium concentrations of the species involved in the reaction can be expressed as:

[SO₂] = (2.60 - x) mol/L

[NO₂] = (x) mol/L

[SO] = (1.20) mol/L

[NO] = (x) mol/L

Using the equilibrium constant expression for the given reaction:

Kc = ([SO][NO]) / ([SO₂][NO₂])

Substituting the values of the concentrations at equilibrium and the given value of Kc, we get:

2.60 = [(1.20)(x)] / [(2.60 - x)(x)]

Simplifying the above equation, we get:

3.12 - 2.60x = x²

Rearranging and solving the quadratic equation, we get:

x² + 2.60x - 3.12 = 0

Using the quadratic formula, we get:

x = (-2.60 ± √(2.60² - 4(1)(-3.12))) / (2(1))

x = (-2.60 ± 2.81) / 2

x = -0.105 or x = 2.96

Since the value of x represents the number of moles of NO₂(g) that react with 2.60 mol of SO₂(g), the negative value of x is not meaningful in this context. Therefore, we can conclude that 2.96 mol of NO₂(g) must be added to 2.60 mol of SO₂(g) to form 1.20 mol of SO(g) at equilibrium.

User Vibert
by
7.6k points