In the following reaction, assume that the reaction enthalpy is independent of t

Question

In the following reaction, assume that the reaction enthalpy is independent of temperature, and the data given below are at 25°C. Calculate the standard reaction Gibbs energy and reaction enthalpy (kJ/mol) and the equilibrium constant at 298K and 450K.

(a) rG°(298K)=___________

(b)K(298K)=___________

(c)rH°(298K)=___________

(d)rG°(450K)=___________

(e)K(450K)=_____________

PbO(s)    +    CO(g)      =       Pb(s)   +    CO2(g).

fG°(KJmol-1) -188.93 -137.17 0               -394.36     

fH°(KJmol-1) -218.99 -110.53 0 -393.51

Explanation / Answer

PbO + CO = Pb + C02

we know that

dGo rxn = dGo products - dGo reactants

so

dGo rxn = dGo PB + dGo C02 - dGo PbO - dGo CO

so

dGo rxn = 0 -394.36 + 188.93 + 137.17

dGo rxn = -68.26

so

dGo rxn at 298 K is -68.26 KJ/mol

b)

now

we know that

dGo = -RT lnK

so

-68.26 x 1000 = -8.314 x 298 x lnK

K = 9.23 x 10^11

so

the K value ate 298 K is 9.23 x 10^11

c)

now

dHrxn = dHo products - dHo reactants

so

dHrxn = dHo Pb + dHo C02 - dHo Pbo - dHo C0

so

dHrxn = 0 - 393.51 + 218.99 + 110.53

dHrxn = -63.99

so

dHrxn at 298 K is -63.99 KJ/mol

d)

now

At 298 K

dGo = dHo - TdSO

so

-68.26 x 1000 = -63.99 x 1000 - ( 298 x dSo)

dSo = 14.33 J/mol / K

now

at 450 K

dGo = dHo - TdSO

dGo = ( -63.99 x 1000) - ( 450 x 14.33)

dGO = -70.44

so

dGo at 450 K is -70.44 kJ/mol

e)

now

dGo = -RT lnK

-70.44 x 1000 = -8.314 x 450 x lnK

K = 1.5 x 10^8

so

K value at 450 K is 1.5 x 10^8

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