The equilibrium constant, KC, for the following reaction at 726.85 °C is 2.71 × 102.2SO2(g) + O2(g) ⇌ 2SO3(g)The KP for this reaction at the same temperature is
Question
The equilibrium constant, KC, for the following reaction at 726.85 °C is 2.71 × 102.2SO2(g) + O2(g) ⇌ 2SO3(g)The KP for this reaction at the same temperature is
Solution
The relationship between Kp and Kc is given by the equation:
Kp = Kc(RT)^(Δn)
where:
- R is the ideal gas constant (0.0821 L·atm/(K·mol) for this calculation)
- T is the temperature in Kelvin
- Δn is the change in moles of gas in the reaction (moles of gaseous products - moles of gaseous reactants)
First, convert the temperature from Celsius to Kelvin. The conversion formula is K = °C + 273.15, so:
T = 726.85°C + 273.15 = 1000 K
The reaction is 2SO2(g) + O2(g) ⇌ 2SO3(g), so:
Δn = (moles of SO3) - (moles of SO2 + moles of O2) Δn = (2) - (2 + 1) Δn = -1
Now we can substitute these values into the equation:
Kp = Kc(RT)^(Δn) Kp = 2.71 x 10^2 * (0.0821 * 1000)^(-1) Kp = 2.71 x 10^2 / 82.1 Kp = 3.30
So, the Kp for this reaction at the same temperature is 3.30.
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