Question

The equilibrium constant Kc for the reaction is 3.8 10-5 at 727°C. What are Kc and Kp for the following equilibrium at the same temperature?

2 I (g) <=> I2 (g)

Answer 1

`K_p`=`3.12*10^-3`

Explanation:

The relationship between `K_c` and `K_p` is in the formula `K_c=K_p*(RT)^(Delta n)`
The R value used is in atmospheres and the T value is in kelvins. `Delta n` is the sum of the coefficients of products minus the sum of the coefficients of the reactants.
Example
aA+bB`<=>` cC+bB
`Delta n` = `(c+b)-(a+b)`
The `Delta n` for the chemical formula gives is -1 so the product of RT is raised to the -1.
`K_c/((RT)^-1)=K_p`
`(3.8*10^-5)/((.0821* 964.15)^-1)~~3.12*10^-3`