In a 100 L vessel,3 moles of nitrogen and 3 | vedclass

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(B) The reaction for the dissociation of $PCl_5$ is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $PCl_5 = 3$,$N_2 = 3$,$PCl_3 = 0$

Vedclass · Accepted Answer

$66.6$

Vedclass · Answer

(B) The reaction for the dissociation of $PCl_5$ is: $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $PCl_5 = 3$,$N_2 = 3$,$PCl_3 = 0$,$Cl_2 = 0$.Let $\alpha$ be the degree of dissociation of $PCl_5$. At equilibrium,moles are: $PCl_5 = 3(1-\alpha)$,$PCl_3 = 3\alpha$,$Cl_2 = 3\alpha$,$N_2 = 3$.Total moles at equilibrium $n_{total} = 3(1-\alpha) + 3\alpha + 3\alpha + 3 = 6 + 3\alpha$.Using the ideal gas law $PV = nRT$:$3.28 	imes 100 = (6 + 3\alpha) 	imes 0.0821 	imes 500$.$328 = (6 + 3\alpha) 	imes 41.05$.$6 + 3\alpha = \frac{328}{41.05} \approx 7.99$.$3\alpha = 7.99 - 6 = 1.99$.$\alpha = \frac{1.99}{3} \approx 0.663$.Percentage dissociation = $0.663 	imes 100 = 66.3 \% \approx 66.6 \%$.

In a $100 \ L$ vessel,$3 \ moles$ of nitrogen and $3 \ moles$ of $PCl_5$ are taken and heated to $500 \ K$. The equilibrium pressure is $3.28 \ atm$. The percentage degree of dissociation of $PCl_5$ is : (Assume ideal behaviour for all gases).

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