Using tabulated thermodynamic data from Appendix D, compute (a) K eq at 298 K; (

Question

Using tabulated thermodynamic data from Appendix D, compute (a) Keq at 298 K; (b) the temperature at which the equilibrium pressure is 1.00 bar; and (c) Keq at 603 K for the following reaction:

Hg(g) + HgCl2(s) Hg2Cl2(s)
(a) Keq at 298 K.

(b) the temperature at which the equilibrium pressure is 1.00 bar.

(c) Keq at 603 K

For (a) I tried to use Keq = e^-G/RT but wasnt able to get the right answer

e^-(210.745kJ/mol / ((8.3144598*10^-3) kJ/K*mol * 298K ) = 0
Im not sure how to get (b) and (c)

HgCl2(s)               H(kJ/mol)= -224.3         G (kJ/mol rxn )= -178.6 S(J/mol ·K) = 146.0

Hg2Cl2(s)             H(kJ/mol)= -265.22 G (kJ/mol rxn )= -210.745 S(J/mol ·K) = 192.5

H2(g)     S=130.68 J/mol K

Explanation / Answer

for the reaction HgCl2(s)+Hg(g)<-------->Hg2Cl2(s)

deltaG= 1* deltaG of Hg2Cl2-(1* deltaG of HgCl2+1*deltaG of Hg) , 1,1 and 1 are coefficients of Hg2Cl2, HgCl2 and Hg respectively.

for Hg(g) at standard state, deltaG=0

deltaG0 = -210.745-(-178.6)= -32.145 Kj/mole

deltaG0=-RT lnK

lnK =-deltaGo/RT=32.145*1000/(8.314*298)=12.97

K= 431230

for the reaction on similar grounds used for deltaG calculations,

deltaH= -265.22-(-224.33)=-40.89 kj/mole

deltaS=192.5-(146+130.68)=-84.2 J/mole.K

deltaG= deltaH-T*deltaS

given T= 693K

deltaG= -40.89*1000+84.2*693 =17405.16 J/mole

since deltaG=-RTlnK

lnK= -deltaG/RT= -17405.16/(8.314*693), K=0.048758

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