A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600◦C. At that temperature, the concentration of SO2 is found to be 1.14 mol/L, the concentration of O2 is 1.39 mol/L, and the concentration of SO3 is 3.18 mol/L. Using the balanced chemical equation, find the equilibrium constant for this system.
To find the equilibrium constant (Kc) for the reaction, we need to write the balanced chemical equation for the reaction first:
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
Now, we can determine the expression for the equilibrium constant (Kc) using the concentrations of the reactants and products:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Substituting the given concentrations:
Kc = (3.18 mol/L)^2 / ((1.14 mol/L)^2 * (1.39 mol/L))
Calculating this expression will give us the equilibrium constant value for the system.
To find the equilibrium constant for this reaction, we need to use the balanced chemical equation, as you mentioned. The balanced equation for the reaction is:
2 SO2(g) + O2(g) -> 2 SO3(g)
The equilibrium constant expression for this reaction can be written as:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Here, [SO3], [SO2], and [O2] represent the equilibrium concentrations of sulfur trioxide, sulfur dioxide, and oxygen, respectively.
Given the concentrations of SO2, O2, and SO3:
[SO2] = 1.14 mol/L
[O2] = 1.39 mol/L
[SO3] = 3.18 mol/L
Now we can substitute these values into the equilibrium constant expression:
Kc = (3.18)^2 / ((1.14)^2 * 1.39)
Calculating this expression will give us the value of the equilibrium constant for the system.
Kc = 8.091 / 1.791
Kc ≈ 4.52
Therefore, the equilibrium constant for this system at 600°C is approximately 4.52.