Dissociation of a gas A_2 takes place accordi | vedclass

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(B) The reaction is $A_{2(g)} \rightleftharpoons 2A_{(g)}$.First, calculate the standard Gibbs free energy change for the reaction: $\Delta_r G^{\circ

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(B) The reaction is $A_{2(g)} ightleftharpoons 2A_{(g)}$.First, calculate the standard Gibbs free energy change for the reaction: $\Delta_r G^{\circ} = 2 	imes \Delta G_f^{\circ}(A) - \Delta G_f^{\circ}(A_2) = 2 	imes (-50.832) - (-100.00) = -101.664 + 100.00 = -1.664 \ kJ \ mol^{-1} = -1664 \ J \ mol^{-1}$.Using the relation $\Delta_r G^{\circ} = -RT \ln K_p$:$-1664 = -8.3 	imes 300 	imes \ln K_p$$\ln K_p = \frac{1664}{2490} \approx 0.668 \approx 0.693$ (using the provided $\ln 2 = 0.693$ approximation).Thus, $K_p = 2$.For the dissociation $A_2 ightleftharpoons 2A$, if $\alpha$ is the degree of dissociation, $K_p = \frac{4\alpha^2 P}{(1-\alpha^2)}$.Given $P = 1 \ bar$, $2 = \frac{4\alpha^2}{1-\alpha^2} \implies 2 - 2\alpha^2 = 4\alpha^2 \implies 6\alpha^2 = 2 \implies \alpha^2 = \frac{1}{3} = 0.3333$.$\alpha = (33.33 	imes 10^{-2})^{1/2}$.Comparing with $(x 	imes 10^{-2})^{1/2}$, we get $x = 33$.

Substance	$\Delta G_f^{\circ} / kJ \ mol^{-1}$
$A_2$	$-100.00$
$A$	$-50.832$

List-$X$	List-$Y$
$(A)$ $A_{(g)} \rightleftharpoons B_{(g)} + \text{Heat}$	$(i)$ Equilibrium constant
$(B)$ $r_b/r_f$	$(ii)$ Favored at low temperature
$(C)$ $r_f/r_b$	$(iii)$ [Equilibrium constant]$^{-1}$
$(D)$ $2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$	$(iv)$ $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
$(E)$ Effect of pressure	$(V)$ $\Delta n < 0$

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