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(A) Reaction: $A(g) \rightleftharpoons B(g)$,$K_{c} = 4.0$.Initial moles: $n_{A} = 1, n_{B} = 0, n_{He} = 1$.Moles at equilibrium: $n_{A} = 1-x, n_{B}

Vedclass · Accepted Answer

$3.28, 2.624$

Vedclass · Answer

(A) Reaction: $A(g) ightleftharpoons B(g)$,$K_{c} = 4.0$.Initial moles: $n_{A} = 1, n_{B} = 0, n_{He} = 1$.Moles at equilibrium: $n_{A} = 1-x, n_{B} = x, n_{He} = 1$.Total moles at equilibrium: $n_{total} = (1-x) + x + 1 = 2$.Total pressure $P_{total} = \frac{n_{total}RT}{V} = \frac{2 	imes 0.082 	imes 400}{10} = 6.56 	ext{ atm}$.For the reaction $A(g) ightleftharpoons B(g)$,$\Delta n = 1 - 1 = 0$,so $K_{p} = K_{c} = 4.0$.$K_{p} = \frac{P_{B}}{P_{A}} = \frac{x_{B} 	imes P_{total}}{x_{A} 	imes P_{total}} = \frac{x / 2}{(1-x) / 2} = \frac{x}{1-x} = 4.0$.$x = 4 - 4x \implies 5x = 4 \implies x = 0.8$.Partial pressure of $He$: $P_{He} = \frac{n_{He}}{n_{total}} 	imes P_{total} = \frac{1}{2} 	imes 6.56 = 3.28 	ext{ atm}$.Partial pressure of $B(g)$: $P_{B} = \frac{n_{B}}{n_{total}} 	imes P_{total} = \frac{0.8}{2} 	imes 6.56 = 2.624 	ext{ atm}$.Thus,the partial pressures are $3.28 	ext{ atm}$ and $2.624 	ext{ atm}$ respectively.

List-$I$ (Equations)	List-$II$ (Type of processes)
$A. K_p > Q$	$(i)$ Non-spontaneous
$B. \Delta G^\circ < RT \ln Q$	$(ii)$ Equilibrium
$C. K_p = Q$	$(iii)$ Spontaneous and endothermic
$D. T > \frac{\Delta H}{\Delta S}$	$(iv)$ Spontaneous

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