Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of SO2(g).
2SO2(g)+O2(g)=2SO3(g) Kc=1.7*10^8
[SO3]aq=0.0034 M
[O2]aq=0.0018 M
Surely you don't mean aqueous for (SO3) and (O2).
To determine the equilibrium concentration of SO2(g), we can use the stoichiometry of the balanced equation and the given equilibrium concentrations of SO3(g) and O2(g).
The balanced equation is:
2SO2(g) + O2(g) ⟶ 2SO3(g)
Let's assume the equilibrium concentration of SO2(g) is x M. Since two moles of SO2(g) react to form two moles of SO3(g), the concentration of SO3(g) at equilibrium will be 2x M.
Using the equilibrium constant expression, Kc, we can write:
Kc = [SO3(g)]^2 / ([SO2(g)]^2 * [O2(g)])
Given:
Kc = 1.7 * 10^8
[SO3] = 0.0034 M
[O2] = 0.0018 M
Plugging in the values, we get:
1.7 * 10^8 = (2x)^2 / [(x)^2 * (0.0018)]
Simplifying the equation:
1.7 * 10^8 = 4x^2 / (0.0018x^2)
Cross-multiplying:
1.7 * 10^8 * 0.0018x^2 = 4x^2
3.06 * 10^5 x^2 = 4x^2
3.06 * 10^5 = 4
x^2 = 3.06 * 10^5 / 4
x^2 = 7.65 * 10^4
Taking the square root of both sides to solve for x:
x = √(7.65 * 10^4)
x ≈ 276.68 M
Therefore, the equilibrium concentration of SO2(g) is approximately 276.68 M.