At 1100 K, Kp = 0.25 for the reaction
2SO2(g) + O2(g) equilibrium with 2SO3(g)
What is the value of Kc at this temperature?
Kc = Kp-delta n
is .25 the right answer? if Kp = to .25 and -delta n is the # of mols in prod - # of mols in reactants then that equals out to 1. please tell me if i messed up somewhere
No. You haven't multiplied by RT but I goofed and didn't put that in my answer.
Kc = Kp(RT)-delta n
Kc = 0.25[0.08205*1100)-(-1)
You are right that delta n is 2-3 = -1
To find the value of Kc at a given temperature, you need to use the relationship between Kc and Kp. The equation is:
Kp = Kc(RT)^(Δn)
Where:
- Kp is the equilibrium constant in terms of partial pressures,
- Kc is the equilibrium constant in terms of molar concentrations,
- R is the ideal gas constant (0.0821 L.atm/mol.K),
- T is the temperature in Kelvin,
- Δn is the change in the moles of gas between the reactants and products.
In this case, the reaction is:
2SO2(g) + O2(g) ⇌ 2SO3(g)
Δn = (2 + 0) - (0 + 2) = 0
Given that Kp = 0.25 at 1100 K, we can substitute the values into the equation.
0.25 = Kc(0.0821)(1100)^0
Since Δn = 0, we can simplify the equation to:
0.25 = Kc(0.0821)
Now, we can solve for Kc:
Kc = 0.25 / 0.0821
Kc ≈ 3.05
Therefore, at 1100 K, the value of Kc for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is approximately 3.05.