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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?

Md Ariful Islam

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Md Ariful Islam

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$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:
$N_2=30120481927.710842mol$

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