Question

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

After calculating the expression, you would get that \(\Delta G_f^\circ \approx -289.78 \,kJ \cdot mol^{-1}\). This represents the Gibbs free energy change for the given reaction at \(25°C\).

Step-by-step solution

Convert Celsius to Kelvin

Firstly, we need to convert the temperature from Celsius to the Kelvin scale, because the formula for Gibbs free energy involves absolute temperature. It can be done by applying the formula, \(T(K) = T(°C) + 273.15\). Therefore, for \(T = 25°C\), in Kelvin it is \(298.15K\).

Identify Known Values

Next, recognize all known quantities from the exercise. We have the equilibrium constant \(K_P = 5.62 \times 10^{35}\), the universal gas constant \(R = 8.3145 \, JK^{-1}mol^{-1}\) or \(R = 0.0083145 \, kJ K^{-1} mol^{-1}\) for conciseness, and the temperature \(T = 298.15 K\) from the previous step.

Use the Formula

Now, use the relation \(\Delta G_f^\circ = -RT\ln K_P\) to calculate Gibbs free energy. Plug in the known values and compute the result.

Calculate the Gibbs Free Energy

By using the given equation and values, \(\Delta G_f^\circ = - (0.0083145 \, kJ K^{-1} mol^{-1}) \times (298.15 K) \times \ln(5.62 \times 10^{35})\), perform the calculation to obtain the value of \(\Delta G_f^\circ\).

Key concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K_P \) or \( K_c \), plays a pivotal role in determining the direction and extent of a chemical reaction. It is pivotal for understanding the balance between the reactants and products in a chemical equation. The equilibrium constant for the given reaction \( \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) \) is reported as \(5.62 \times 10^{35}\), indicating a strong tendency for the formation of the product, \( \mathrm{COCl}_{2}(g) \).- **Interpreting the magnitude**: A very large \( K_P \) value signifies that, at equilibrium, the concentration of products greatly exceeds that of reactants. Hence, this reaction heavily favors the production of \( \mathrm{COCl}_{2} \).- **Dependency on temperature**: The equilibrium constant is temperature-dependent. This means any change in temperature could alter \( K_P \), affecting the balance of the equilibrium.Understanding the equilibrium constant helps predict how the system will respond to external changes, such as shifts in temperature.

Temperature Conversion

Converting temperature from Celsius to Kelvin is crucial when dealing with thermodynamic calculations, such as determining Gibbs Free Energy. This is because thermodynamic equations often require the absolute temperature, which is measured in Kelvin.- **Formula for conversion**: To convert from Celsius to Kelvin, use the formula: \[ T(K) = T(°C) + 273.15 \] This ensures that you are working with the correct temperature scale needed for accurate calculations.- **Example**: For \( T = 25^{\circ}C \), the equivalent in Kelvin is \[ 298.15 K \]Understanding and using the Kelvin scale prevents errors in thermodynamic calculations, such as those involving the Gibbs Free Energy.

Universal Gas Constant

The universal gas constant, denoted as \( R \), is a fundamental constant in thermodynamics that appears in various equations, including the formula for Gibbs Free Energy. Its values, depending on the units used, are:- **\(8.3145\, JK^{-1}mol^{-1}\)** or - **\(0.0083145\, kJ K^{-1} mol^{-1}\)**This constant relates energy scales to temperature scales, facilitating the conversion between different forms of energy. In the context of the Gibbs Free Energy calculation, \( R \) helps link the equilibrium constant \( K_P \) with the temperature \( T \), through the formula:\[\Delta G_f^\circ = -RT \ln K_P\]The choice of unit for \( R \) should match the units of the other parameters in the equation to ensure consistency and accuracy. It simplifies complex relationships, making calculations manageable and understandable.