Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
Medium$6.9 \ g$ of $N_2O_4$ is taken in a $0.5 \ L$ closed vessel at $400 \ K$. For the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the total pressure at equilibrium is $9.15 \ atm$. Calculate $K_c$,$K_p$,and the partial pressure of each component.
(A) $1$. Moles of $N_2O_4$ initially: $n = \frac{6.9 \ g}{92 \ g/mol} = 0.075 \ mol$.
$2$. Initial pressure $P_i$ of $N_2O_4$: $P_i = \frac{nRT}{V} = \frac{0.075 \times 0.0821 \times 400}{0.5} = 4.926 \ atm$.
$3$. Let $x$ be the degree of dissociation. At equilibrium: $P_{N_2O_4} = P_i(1-x)$ and $P_{NO_2} = 2P_ix$.
$4$. Total pressure $P_T = P_i(1-x) + 2P_ix = P_i(1+x) = 9.15 \ atm$.
$5$. $1+x = \frac{9.15}{4.926} \approx 1.857$,so $x = 0.857$.
$6$. Partial pressures: $P_{N_2O_4} = 4.926(1-0.857) = 0.704 \ atm$ and $P_{NO_2} = 2 \times 4.926 \times 0.857 = 8.442 \ atm$.
$7$. $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(8.442)^2}{0.704} \approx 101.25 \ atm$.
$8$. $K_c = K_p(RT)^{-\Delta n} = 101.25 \times (0.0821 \times 400)^{-1} = \frac{101.25}{32.84} \approx 3.083 \ mol \ L^{-1}$.
$2$. Initial pressure $P_i$ of $N_2O_4$: $P_i = \frac{nRT}{V} = \frac{0.075 \times 0.0821 \times 400}{0.5} = 4.926 \ atm$.
$3$. Let $x$ be the degree of dissociation. At equilibrium: $P_{N_2O_4} = P_i(1-x)$ and $P_{NO_2} = 2P_ix$.
$4$. Total pressure $P_T = P_i(1-x) + 2P_ix = P_i(1+x) = 9.15 \ atm$.
$5$. $1+x = \frac{9.15}{4.926} \approx 1.857$,so $x = 0.857$.
$6$. Partial pressures: $P_{N_2O_4} = 4.926(1-0.857) = 0.704 \ atm$ and $P_{NO_2} = 2 \times 4.926 \times 0.857 = 8.442 \ atm$.
$7$. $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(8.442)^2}{0.704} \approx 101.25 \ atm$.
$8$. $K_c = K_p(RT)^{-\Delta n} = 101.25 \times (0.0821 \times 400)^{-1} = \frac{101.25}{32.84} \approx 3.083 \ mol \ L^{-1}$.
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