Question

Question #a90d2

Answer 1

For the reaction, 2 N`H_3` (g) <-----> `N_2` (g) + 3 `H_2` (g) at 298 K, `K_c` = 2.8 x `10^ (-9)` What is the value `K_p` of for this reaction?

`K_c` = `(RT)^(∆n)`. `K_p`

Change in number of moles : total number of moles of products - total number of moles of reactants.
∆n = ∑ `n_ p` - ∑ `n_ r`

∆n = (1+3) - 2 = -2

`K_c` = `(RT)^(∆n)`. `K_p`

`K_p` = `(RT)^(∆n)` / `K_c`

`K_p` = `(0.08206 X 1338)^(-2)` / 2.8 x `10^ (-9)`
= 1 / `(0.08206 X 1338)^(2)` x 2.8 x `10^ (-9)`

`K_p` = 1/ 33755 x `10^ (-9)` = 29626.