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Chapter 17: Problem 30

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The standard Gibbs Free Energy change, \(\Delta G_{f}^{\circ}\), for the formation of COCl2 at 25°C is -231.36 kJ/mol.

Step by step solution

01

Convert the temperature from Celsius to Kelvin

The temperature given in Celsius needs to be converted to Kelvin for the calculation. This can be done using the equation `T(K) = T(°C) + 273.15`. So, T(25°C) = 25 + 273.15 = 298.15 K.
02

Plug in the values into the Gibbs free energy equation

Now, the Gibbs free energy change can be calculated using the formula `ΔG = -RT ln K`, where R is 0.008314 kJ/mol.K, T is 298.15 K and K is 5.62 × 10^{35}, Plugging in the values gives `ΔG = -(0.008314 kJ/mol.K * 298.15 K * ln(5.62 × 10^{35}))`.
03

Calculate ΔG

On calculating the above, we get an approximate value for `ΔG = -231.36 kJ/mol`. Hence, \(\Delta G_{f}^{\circ}\) for COCl2 at 25°C is -231.36 kJ/mol.

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Most popular questions from this chapter

For a reaction with a negative \(\Delta G^{\circ}\) value, which of the following statements is false? (a) The equilibrium constant \(K\) is greater than one. (b) The reaction is spontaneous when all the reactants and products are in their standard states. (c) The reaction is always exothermic.

Explain the following nursery rhyme in terms of the second law of thermodynamics. Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King's horses and all the King's men Couldn't put Humpty together again.

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