Problem 120 - Chemistry: An Atoms-Focused Approach

Question

The value of \(K_{e}\) for the reaction \(A \rightleftharpoons B\) is 0.455 at \(50^{\circ} \mathrm{C}\) and 0.655 at \(100^{\circ} \mathrm{C}\). Calculate \(\Delta H^{*}\) for the reaction.

Step-by-Step Solution

Answer

Question: Calculate the change in enthalpy (ΔH*) for the reaction A ⇌ B given their equilibrium constants (Ke) at temperatures 50°C (0.455) and 100°C (0.655). Answer: The change in enthalpy (ΔH*) for the reaction A ⇌ B is approximately -2858.52 J/mol.

Step 1: Convert temperatures to Kelvin

To use the Van't Hoff equation, we need the temperature in Kelvins (K), so we need to convert the given temperatures from Celsius to Kelvin by adding 273.15. Temperature 1 (T1) = 50°C + 273.15 = 323.15 K Temperature 2 (T2) = 100°C + 273.15 = 373.15 K

Step 2: Write down the Van't Hoff equation

The Van't Hoff equation relates equilibrium constant, the change in enthalpy, and temperature. The equation is given by: \( \frac{-\Delta H^*}{R} = \frac{\ln K_{e2} - \ln K_{e1}}{\frac{1}{T_2} - \frac{1}{T_1}} \) Where ΔH* = change in enthalpy, R = gas constant (8.314 J/mol·K), \(K_{e1}\) and \(K_{e2}\) are the equilibrium constants at temperatures T1 = 323.15 K and T2 = 373.15 K, respectively.

Step 3: Calculate the logarithms of equilibrium constants

Now, let's calculate the natural logarithm (ln) of both equilibrium constants. \(\ln K_{e1}\) = \(\ln (0.455)\) \(\ln K_{e2}\) = \(\ln (0.655)\)

Step 4: Substitute the values into the Van't Hoff equation

Now, let's substitute our values \(\ln K_{e1}\), \(\ln K_{e2}\), \(T_1\), and \(T_2\) into the Van't Hoff equation. \( \frac{-\Delta H^*}{8.314} = \frac{\ln (0.655) - \ln (0.455)}{\frac{1}{373.15} - \frac{1}{323.15}} \)

Step 5: Solve the equation for ΔH*

Now we have all the values needed to solve for ΔH*. Perform the calculations and isolate the ΔH* term on the left side. \(\Delta H^* = -8.314 \times (\frac{\ln (0.655) - \ln (0.455)}{\frac{1}{373.15} - \frac{1}{323.15}}) \) After calculating the result, we get: \(\Delta H^* ≈ -2858.52 \thinspace J \thinspace mol^{-1}\)

Step 6: Conclusion

The change in enthalpy (ΔH*) for the reaction A ⇌ B is approximately -2858.52 J/mol.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_e \), is a crucial concept in chemical equilibrium. It is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. For the reaction \( A \rightleftharpoons B \), \( K_e \) is calculated by dividing the concentration of \( B \) by the concentration of \( A \).
The value of the equilibrium constant helps determine the directionality of a reaction under constant temperature and pressure.

• If \( K_e \) is greater than 1, the products are favored at equilibrium.
• If \( K_e \) is less than 1, the reactants are favored.

In the Van't Hoff equation, the equilibrium constants at two different temperatures allow us to calculate the change in enthalpy, \( \Delta H^* \). The equation uses the natural logarithm of \( K_e \) values to determine how \( \Delta H^* \) shifts with temperature.

Change in Enthalpy (ΔH)

The change in enthalpy, \( \Delta H^* \), is a measure of the heat absorbed or released in a chemical reaction under constant pressure.
It signifies the energy change as products form from reactants. If \( \Delta H^* \) is negative, the reaction is exothermic, releasing heat. Conversely, a positive \( \Delta H^* \) indicates an endothermic process, absorbing heat.
To calculate \( \Delta H^* \) using the Van’t Hoff equation, we set up the formula:
\[\frac{-\Delta H^*}{R} = \frac{\ln K_{e2} - \ln K_{e1}}{\frac{1}{T_2} - \frac{1}{T_1}}\]
where \( R \) is the gas constant. By finding the difference in natural logs of the equilibrium constants and the reciprocal temperatures, we ascertain \( \Delta H^* \). This process provides insights into the thermodynamic properties of reactions, helping predict how changes in temperature impact reaction feasibility.

Temperature Conversion to Kelvin

Temperature conversion to Kelvin is a fundamental step in computational chemistry, especially when applying the Van’t Hoff equation.
The Kelvin scale is the absolute temperature scale used in scientific calculations, which prevents errors due to negative values inherent to Celsius or Fahrenheit.

• To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
  For example:
• 50°C becomes \( 50 + 273.15 = 323.15 \) K
• 100°C becomes \( 100 + 273.15 = 373.15 \) K

Using Kelvin ensures uniformity in calculations, particularly in thermodynamics, where temperature has a direct relation to molecular kinetic energy and behavior.
In the Van't Hoff context, accurate temperature conversion aids in correctly solving for thermal properties such as \( \Delta H^* \), influencing our understanding of equilibrium dynamics.