# 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?

Mar 6, 2017

${N}_{2} = 30120481927.710842 m o l$

#### Explanation:

$K c = {\left[N {O}_{2}\right]}^{2} / \left(\left[N 2\right] {\left[O 2\right]}^{2}\right)$

The Kc is measured at the equilibrum

$8.3 \cdot {10}^{-} 10 = {x}^{2} / \left(\left[{N}_{2}\right] {\left[x\right]}^{2}\right)$

Conc of ${N}_{2} \text{when} N {O}_{2} = {O}_{2}$ = 1204819277.108434mol

$8.3 \cdot {10}^{-} 10 = \frac{5 {x}^{2}}{{N}_{2} \cdot {x}^{2}}$

Let's solve for ${N}_{2}$.

8.3(10^−10)= (5x)^2 /(x^2N_2)

Step 1: Multiply both sides by ${N}_{2}$

$0 {N}_{2} = 25$

Step 2: Divide both sides by 0.

$\frac{0 {N}_{2}}{0} = \frac{25}{0}$

${N}_{2} = 30120481927.710842 m o l$

${N}_{2} = 30120481927.710842 m o l$