Question

Using the following thermochemical tables, calculate the indicated equilibrium constant at 298 K for the following reactions. (a) H2(g) + I2(s) -> 2HI(g) Kp = ? (b) NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq) Kb = ? (c) Fe(OH)2(s) = Fe2+(aq) + 2OH-(aq) Ksp = ?

Substance ΔHo (kJ/mol) ΔGo (kJ/mol) So (J/mol K)

Al(s)             0                          0                   28.32

Al2O3(s)    -1669.8             -1576.5               51.0

CH3OH(l) -238.6               -166.23              126.8

CO(g)        -110.5               -137.2                197.9

CO2(g)       -393.5               -394.4                213.6

Cl2(g)            0                         0                   222.96

Cl –(aq)       -167.2               -131.2               56.5

Fe2+(aq)       -87.86              -84.93               113.4

Fe3+(aq)       -47.69              -10.54               293.3

Fe(OH)2(s) -568.2             -483.6                79.5

HCl(g)         -92.30             -95.27               186.69

HI(g)            25.94               1.30                 206.3

OH – (aq)     -230.0             -157.3              -10.7

H2(g)               0                       0                  130.58

H2CO3(aq)    -699.65          -623.09             187.4

H2O(g)          -241.82          -228.57            188.83

H2O(l)           -285.83          -237.13            69.91

NH3(g)          -46.19            -16.66              192.5

NH3(aq)        -80.29            -26.50              111.3

NH4Cl(s)       -314.4           -203.0               94.6

NH4+(aq)       -132.5           -79.31               113.4

Answer 1

Solution :-

Part a) H2(g) + I2(s) ---- > 2 HI(g)

Using the standard free energy of formation we can find the free energy change of the reaction

Delta Grxn = [sum of delta Go product] –[ sum of delta Go reactant ]

                    = [HI*2] – [(H2*1)+(I2*1)]

                     = [1.30 *2] – [(0*1)+(0*1)]

                     = 2.60 kJ/mol

2.60kJ per mol * 1000 J / 1 kJ= 2600 J/mol

Using the delta G we can find the kp

Delta Go = - RT ln Kp

Kp = e^(delta Go / - RT)

Kp=e^(2600 J per mol / (-8.314J per mol K * 298 K))

Kp= 0.350

Part b) NH3(aq) + H2O(l)   ---- > NH4^+(aq) + OH^-(aq)

Delta Grxn = [sum of delta Go product] –[ sum of delta Go reactant ]

                    = [(NH4^+ * 1)+(OH^- * 1)] – [(NH3*1)+(H2O*1)]

                    =[(-79.31*1)+(-157.3*1)]-[(-26.50*1)+(-237.13*1)]

                    = 27.02 kJ/mol

27.02kJ per mol * 1000 J / 1 kJ= 27020 J/mol

Delta Go = - RT ln Kb

Kb = e^(delta Go / - RT)

Kb=e^(27020 J per mol / (-8.314J per mol K * 298 K))

Kb= 1.84*10^-5

Part c)

Fe(OH)2(s) ---- > Fe^2+(aq) + 2OH^-(aq)

Delta Grxn = [sum of delta Go product] –[ sum of delta Go reactant ]

                    =[(Fe^2+ * 1)+(OH^- * 2)] – [Fe(OH)2 *1]

                   = [(-84.93*1)+(-157.3*2)]-[-483.6*1]

                   = 84.07 kJ/mol

84.07 kJ per mol * 1000 J / 1 kJ = 84070 J/mol

Delta Go = - RT ln Ksp

Ksp = e^(delta Go / - RT)

Ksp=e^(84070 J per mol / (-8.314J per mol K * 298 K))

Ksp= 1.83*10^-15