Question

From the information in the Data section of the textbook, calculate the equilibrium constant at 338 K for the reaction: assuming that the reaction enthalpy is independent of temperature. Answer: 0.0835 The degree of dissociation, a, for the following reaction is 0.655 at 298 K and 1.00 bar total pressure. Find K Answer: 300 Given K 9.18E-8 for the reaction: laqi and the standard Gibbs energy of formation of AB2(s) is-152.4 kJ mol- find the standard Gibbs energy of formation of AB2(aq), also in kJ mol-1 Answer: -192.48 MacBook
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can you only do uestion 2 and 3 for me

step by step

Answer 1

2. Reaction is:

	N2O4(g)	<-->	2NO2(g)
Initial (bar)	1		0
Change (bar)	-$\alpha$		+2$\alpha$
Equilibrium (bar)	1-$\alpha$		2$\alpha$

K_p = (P_No2)² / P_N2O4

P_N2O4 = x_n2O4 P = [(1-$\alpha$) / (1+$\alpha$)]P

and P_NO2 = x_NO2 P = [2$\alpha$/(1+$\alpha$)] P

K_p = ([2$\alpha$/(1+$\alpha$)] P)² / ([(1-$\alpha$) / (1+$\alpha$)]P) = 4$\alpha$²P / (1-$\alpha$²)

K = 4(0.655)²(1 bar) / (1-(0.655)²) = 3.00 [not 300]

3. As we know

$\Delta$G = -RT ln K_c = -(8.314 J /K.mol * 298.15 K) ln (9.18*10^-8) = -40165.93 J/mol = -40.166 kJ/mol

Now,$\Delta$G_rxn = $\Delta$G products - $\Delta$G_reactants

-40.166 kJ/mol = $\Delta$G(AB₂,(aq) ) - $\Delta$G(AB₂(s))

-40.166 kJ/mol = $\Delta$G(AB₂,(aq) ) - (-152.4kJ/mol)

$\Delta$G(AB₂,(aq) ) = 112.2 kJ/mol