For the reaction: $N_{2} (g) + 2 O_{2} (s) ⇌ 2 N O_{2} (g)$ $N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)$ , $K_{c}$ $K_c$ - $8.3 x 10^{- 10}$ $8.3x10^-10$ at 25 °C. What is the concentration of $N_{2}$ $N_2$ gas at equilibrium when the concentration of $N O_{2}$ $NO_2$ is five times the concentration of $O_{2}$ $O_2$ gas?

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Answer 1 · 2017-03-06T18:04:38

$N_{2} = 30120481927.710842 m o l$ $N_2=30120481927.710842mol$

$K c = \frac{{[N O_{2}]}_{2}^{2}}{[N 2] {[O 2]}^{2}}$ $Kc = [NO_2]^2/ ([N2][O2]^2)$

The Kc is measured at the equilibrum

$8.3 \cdot 10^{- 10} = \frac{x^{2}}{[N_{2}] {[x]}^{2}}$ $8.3 * 10^-10 = x^2/([N_2][x]^2)$

Conc of $N_{2} when N O_{2} = O_{2}$ $N_2 "when" NO_2 = O_2$ = 1204819277.108434mol

$8.3 \cdot 10^{- 10} = \frac{5 x^{2}}{N_{2} \cdot x^{2}}$ $8.3 * 10^-10 = (5x^2)/(N_2 * x^2)$

Let's solve for $N_{2}$ $N_2$ .

$8.3 (10^{-} 10) = \frac{{(5 x)}^{2}}{x^{2} N_{2}}$ $8.3(10^−10)= (5x)^2 /(x^2N_2)$

Step 1: Multiply both sides by $N_{2}$ $N_2$

$0 N_{2} = 25$ $0N_2=25$

Step 2: Divide both sides by 0.

$\frac{0 N_{2}}{0} = \frac{25}{0}$ $(0N_2)/0 = 25/0$

$N_{2} = 30120481927.710842 m o l$ $N_2=30120481927.710842mol$

Answer:
$N_{2} = 30120481927.710842 m o l$ $N_2=30120481927.710842mol$

For the reaction: N2(g)+2O2(s)⇌2NO2(g)N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g), KcK_c - 8.3x10−108.3x10^-10 at 25 °C. What is the concentration of N2N_2 gas at equilibrium when the concentration of NO2NO_2 is five times the concentration of O2O_2 gas?