(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Using information from tables, calculate the equilibrium constant K of the following reaction at 32 degrees C.

[tex]PCl_5 \rightarrow PCl_3 + Cl_2[/tex]

[tex]\Delta H^0_{rxn} = 68.6kJ[/tex]

a) [tex]4.62 \times 10^{-3}[/tex]atm

b) 216 atm

c) [tex]7.81 \times 10^{-11}[/tex]atm

d) 163 atm

2. Relevant equations

[tex] \Delta G^0 = -RT\ln{K_p}[/tex]

[tex] \Delta G^0 = \Delta H^0 - T\Delta S^0[/tex]

3. The attempt at a solution

First attempt:

Calculated the change in Gibbs energy using the energy of formation of the compounds, then plugged into the first equation to get Kp. Wrong answer.

[tex] \Delta G^0 = 0 + (-267.8\frac{kJ}{mol}) - (-305\frac{kJ}{mol}) = 37.2kJ[/tex]

[tex]K_p = e^{\frac{-37200J}{8.3145(305K)}} = 4.289 \times 10^{-7}[/tex]

Second attempt:

Calculated the change in entropy using values from tables, plugged into 2nd equation to get change in Gibbs, then plugged into first equation to get Kp. Wrong answer.

[tex] \Delta S^0 = 223.08\frac{J}{Kmol} + 311.78\frac{J}{Kmol} - 364.58\frac{J}{Kmol} = 170.28\frac{J}{K}[/tex]

[tex]\Delta G^0 = 68600J - (305K)(170.28\frac{J}{K}) = 16664.6J[/tex]

[tex]\Delta K_p = 1.4 \times 10^{-3}[/tex]

As usual, I'm going nuts. Any help appreciated.

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# Homework Help: Equilibrium constant from thermodynamic values

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