Question

What is Kp?

For the reaction
2A(g)+2B(g)⇌C(g)

Kc = 34.8 at a temperature of 19 ∘C .

Answer 1

`Kp = 0.00253`

Explanation:

`Kp = Kc(RT)^(∆n)`

Where:
`Kp` is the equilibrium constant for partial pressure
`Kc` is the equilibrium constant for concentration
`R = 0.08206 "L*atm"/"mol*K"` (it's a constant)
`T` is the temperature (in Kelvin)
`∆n` is the change in the number of mols (number of mols of products in balanced reaction equation - number of mols in of reactants in balanced reaction equation)

We know that
`Kp` is unknown
`Kc` = 34.8
`R = 0.08206 "L*atm"/"mol*K"`
#T = 19˚C + 273.15 = 292.15 "K" ( this is how you convert from C˚ to K) #∆n = 1 "mol" - 4 "mol" = -3 "mol"#

Plugging in the information we then have:

`Kp = 34.8(0.08206 "L*atm"/"mol*K"*292.15"K")^(-3"mol")`

`Kp = 0.00253`