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Chapter 18: Problem 27

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{f}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$. (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\).

Short Answer

Expert verified
The calculated standard Gibbs free energy change \(\Delta G^\circ\) is 39 kJ/mol. The equilibrium constant \(K_P\) can be calculated using the formula and values given above. The Gibbs free energy change \(\Delta G\) under the given conditions can be calculated using the calculated \(\Delta G^\circ\) and \(K_p\) values along with given pressures.

Step by step solution

01

Calculate Gibbs Free Energy Change (\(\Delta G^\circ\))

The change in Gibbs free energy for a reaction, \(\Delta G^\circ\), can be calculated with the formula \(\Delta G^\circ = \Delta G_f^\circ(\text{products}) - \Delta G_f^\circ(\text{reactants})\). The given \(\Delta G_f^\circ\) values are: \(\mathrm{Cl}_{2}(g)\) is 0, \(\mathrm{PCl}_{3}(g)\) is -286 kJ/mol and \(\mathrm{PCl}_{5}(g)\) is -325 kJ/mol. So, \(\Delta G^\circ = [\Delta G_f^\circ(\mathrm{PCl}_{3}(g)) + \Delta G_f^\circ(\mathrm{Cl}_{2}(g))] - \Delta G_f^\circ(\mathrm{PCl}_{5}(g)) = ([-286 + 0] - (-325)) \, \mathrm{kJ/mol} = 39 \, \mathrm{kJ/mol}\)
02

Convert Gibbs Energy Change from kJ to J

Since the standard gas constant R will be in J, for consistency we need to convert \(\Delta G^\circ\) from kJ to J. So, \(\Delta G^\circ = 39 \times 10^3 \, \mathrm{J/mol}\)
03

Calculate Equilibrium Constant (\(K_P\))

Using the formula \(K_P = \exp{(-\Delta G^\circ/RT)}\) where R is the gas constant (\(8.314 \, \mathrm{J/mol.K}\)) and T is the temperature in Kelvin (298K, as \(25^\circ \mathrm{C}\) is equivalent to 298K), we can find \(K_P = \exp{(-39000/8.314 \times 298)}\). Compute this expression to find the value of \(K_P\)
04

Calculate Gibbs Free Energy Change (\(\Delta G\)) under Given Conditions

Next, calculate the Gibbs free energy change (\(\Delta G\)) at given pressures using the expression \( \Delta G = \Delta G^\circ + RT \ln \left( \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} \right)\). Substitute \( \Delta G^\circ = 39000 \, \mathrm{J/mol}\), the given pressures \(P_{PCl_5}=0.0029 \, \mathrm{atm}\), \(P_{PCl_3}=0.27 \, \mathrm{atm}\), \(P_{Cl_2}=0.40 \, \mathrm{atm}\) and the above mentioned values of R and T to get the value of \(\Delta G\)

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Most popular questions from this chapter

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