Answer:
[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
Answer:
[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
