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. Please enter your answers for rG° (kJ/mol) and rH° (kJ/mol) with 2 decimals. (a) rG°(298K)=____ (b)K(298K)=____(c)rH°(298K)=_____(d)rG°(450K)=____(e)K(450K)=____

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

                                       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

a)

dGRxn at 298k = G-products - G-reactnants= (0 + -394.36) -(-188.93-137.17) = -68.26 kJ/mol

b)

K = 298K

dG° = -RT*ln(K)

K = exp(-dG/(RT)) = exp(6826/(8.312*298)) = 15.7333

c)

dHRxn = Hprod - Hreact = (0 - 393.51) - (-218.99 - 110.53) = -63.99 kJ/mol

e)

K at 450K

ln(k2/k1) = H/*R (1/T1-1/T2)

ln(K2/15.7333) = -63990/8.314 * (1/298 - 1/450)

K2 = 15.7333*exp( -63990/8.314 * (1/298 - 1/450))

K2 = 0.002558

so..

d)

dG at 450 K...

dG = dG° + RT*lnK

dG = -68260 + 8.31*298*ln(0.002558) = -83040.34727 J/mol = -83.04 kJ/mol

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