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(A) Let the initial moles of $B_2$ be $1 \ mol$. At equilibrium,let the degree of dissociation be $\alpha$. The reaction is $B_{2(g)} \rightleftharpoo

Vedclass · Accepted Answer

$0.05$

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(A) Let the initial moles of $B_2$ be $1 \ mol$. At equilibrium,let the degree of dissociation be $\alpha$. The reaction is $B_{2(g)} ightleftharpoons 2B_{(g)}$. Initial moles: $1$ for $B_2$,$0$ for $B$. Equilibrium moles: $(1-\alpha)$ for $B_2$,$2\alpha$ for $B$. Total moles at equilibrium = $(1-\alpha) + 2\alpha = 1+\alpha$. The volume fraction of $B$ atoms is given as $20 \%$,which is equal to the mole fraction of $B$ atoms in the gas mixture. Mole fraction of $B$ = $\frac{2\alpha}{1+\alpha} = 0.2$. $2\alpha = 0.2 + 0.2\alpha \implies 1.8\alpha = 0.2 \implies \alpha = \frac{0.2}{1.8} = \frac{1}{9}$. Partial pressure of $B$ $(P_B)$ = $\frac{2\alpha}{1+\alpha} 	imes P_{total} = 0.2 	imes 1 \ bar = 0.2 \ bar$. Partial pressure of $B_2$ $(P_{B_2})$ = $\frac{1-\alpha}{1+\alpha} 	imes P_{total} = \frac{1 - 1/9}{1 + 1/9} 	imes 1 = \frac{8/9}{10/9} = 0.8 \ bar$. $K_p = \frac{(P_B)^2}{P_{B_2}} = \frac{(0.2)^2}{0.8} = \frac{0.04}{0.8} = 0.05$.

The following equilibrium is established at $STP$. $B_{2(g)} \rightleftharpoons 2B_{(g)}$. Atoms of $B$ occupy $20 \%$ of total volume at $STP$. The total pressure of the system is $1 \ bar$. What is its $K_p$?

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Equilibrium constants $K_1$ and $K_2$ for the following equilibria:
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