At 950K for the reaction 2SO3 ⇌ 2 SO2 + O2, the initial pressure of SO3 is 32 atm. And the
equilibrium pressure of O2 is 5.6 atm. Calculate Kp and Kc.
....................... 2SO3 ⇌ 2 SO2 + O2
I.......................32 atm........0..........0
C.......................-2x.............2x........x
E.....................32-2x............2x........x
The problem tells you that x = 5.6 atm so 2x must be 11.2 and 32-2x must be 32-11.2 = 20.8. Substitute these numbers into the Kp expression and solve.
Kp = p^2(SO2)*pO2/p*2(SO3).
For Kc, remember Kp =Kc*(RT)^delta n Post your work if you get stuck.
To calculate the equilibrium constant Kp and Kc for the given reaction, we need to determine the equilibrium concentrations of the involved species.
Given:
- Initial pressure of SO3 = 32 atm
- Equilibrium pressure of O2 = 5.6 atm
First, let's assume that the equilibrium pressure of SO2 is P (we don't have this value yet). Since the stoichiometry of the reaction is 2:1:1 (2 moles of SO3 produce 2 moles of SO2 and 1 mole of O2), the equilibrium pressure of SO2 will also be P. And the equilibrium pressure of O2 is given as 5.6 atm.
Now, we can set up the equilibrium expression for Kp and Kc:
Kp = (P(SO2))^2 x (P(O2))
Kc = ([SO2]^2 x [O2]) / ([SO3]^2)
To find the value of P (equilibrium pressure of SO2), we will use the ideal gas law: PV = nRT.
Given:
R = 0.0821 L·atm/(mol·K) (Universal gas constant)
T = temperature is not provided
Since we don't have the temperature, we cannot solve for the number of moles (n). So, we need additional information to proceed with finding the equilibrium constant.
Please provide the temperature of the reaction or any additional information required to calculate the equilibrium constants Kp and Kc.