Question

For the reaction: `N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)`, `K_c` - `8.3x10^-10` at 25 °C. What is the concentration of `N_2` gas at equilibrium when the concentration of `NO_2` is five times the concentration of `O_2` gas?

Answer 1

`N_2=30120481927.710842mol`

Explanation:

`Kc = [NO_2]^2/ ([N2][O2]^2)`

The Kc is measured at the equilibrum

`8.3 * 10^-10 = x^2/([N_2][x]^2)`

Conc of `N_2 "when" NO_2 = O_2` = 1204819277.108434mol

`8.3 * 10^-10 = (5x^2)/(N_2 * x^2)`

Let's solve for `N_2`.

`8.3(10^−10)= (5x)^2 /(x^2N_2)`

Step 1: Multiply both sides by `N_2`

`0N_2=25`

Step 2: Divide both sides by 0.

`(0N_2)/0 = 25/0`

`N_2=30120481927.710842mol`

Answer:
`N_2=30120481927.710842mol`