Ammonia under a pressure of 15 atm at 27 ^{ | vedclass

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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 equilibr

Vedclass · Accepted Answer

$61.3$

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 of $a$ moles of $NH_3$ at $27 \ ^{\circ}C$ $(300 \ K)$ is $15 \ atm$.Let $p$ be the pressure of $a$ moles of $NH_3$ at $347 \ ^{\circ}C$ $(620 \ K)$.Using $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (at constant volume),$\frac{15}{300} = \frac{p}{620}$,which gives $p = 31 \ atm$.Since $P \propto n$ at constant $V$ and $T$,the initial pressure $P_i = 31 \ atm$ corresponds to $a$ moles.The final pressure $P_f = 50 \ atm$ corresponds to $(a + 2x)$ moles.Therefore,$\frac{a + 2x}{a} = \frac{50}{31}$.$1 + \frac{2x}{a} = \frac{50}{31} \implies \frac{2x}{a} = \frac{50}{31} - 1 = \frac{19}{31}$.Percentage of $NH_3$ decomposed $= \frac{2x}{a} 	imes 100 = \frac{19}{31} 	imes 100 \approx 61.29 \% \approx 61.3 \%$.

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