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(B) The reaction is $2NH_3 \rightleftharpoons N_2 + 3H_2$. Initial moles: $a, 0, 0$. Moles at equilibrium: $(a-2x), x, 3x$. Total moles at equilibrium

Vedclass · Accepted Answer

$61.33$

Vedclass · Answer

(B) The reaction is $2NH_3 ightleftharpoons N_2 + 3H_2$. Initial moles: $a, 0, 0$. Moles at equilibrium: $(a-2x), x, 3x$. Total moles at equilibrium = $(a-2x) + x + 3x = a + 2x$. Initial pressure $P_1 = 15 \ atm$ at $T_1 = 300 \ K$. Pressure $P_2$ at $T_2 = 620 \ K$ (before dissociation) is calculated using $\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{15}{300} = \frac{P_2}{620} \implies P_2 = 31 \ atm$. Since $PV = nRT$,at constant volume and temperature,$P \propto n$. Initial pressure $P_{initial} = 31 \ atm \propto a$. Final pressure $P_{final} = 50 \ atm \propto (a + 2x)$. Therefore,$\frac{a + 2x}{a} = \frac{50}{31} \implies 1 + \frac{2x}{a} = \frac{50}{31} \implies \frac{2x}{a} = \frac{50}{31} - 1 = \frac{19}{31}$. Percentage dissociation $= \frac{2x}{a} 	imes 100 = \frac{19}{31} 	imes 100 \approx 61.29\% \approx 61.33\%$ (based on provided options).

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