Question

The following equilibrium constants have been determined for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) \(K_{\mathrm{p}}=50.0\) at \(448^{\circ} \mathrm{C}\) and 66.9 at \(350^{\circ} \mathrm{C} .\) Use these data to estimate \(\Delta H^{\circ}\) for the reaction.

Short answer

The change in enthalpy for the reaction (\(\Delta H^{\circ}\)) is estimated to be -3.27 kJ/mol.

Step-by-step solution

Gather information

The Van't Hoff equation can be re-arranged as follows: \(\ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^{\circ}}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \). From the exercise, the following information is given: \(K_1 = 50.0\), \(T_1 = 448^{\circ}C = 448 + 273.15 = 721.15 K\) (as the temperature must be in Kelvin for this equation), \(K_2 = 66.9\), and \(T_2 = 350^{\circ}C = 350 + 273.15 = 623.15 K\). \(R = 8.3145 \, J \, K^{-1} \, mol^{-1}\), the ideal gas constant.

Substitute the values into the Van't Hoff equation

Substitute the above values into the Van't Hoff equation: \(\ln \left(\frac{66.9}{50.0}\right) = -\frac{\Delta H^{\circ}}{8.3145} \times \left(\frac{1}{623.15} - \frac{1}{721.15}\right)\). This can be further simplified: \(\ln(1.338) = -\frac{\Delta H^{\circ}}{8.3145} \times (-0.000701)\). Calculating the value inside the logarithm gives approximately 0.2874.

Solve for the change in enthalpy

We then solve for \(\Delta H^{\circ}\) by multiplying both sides by -8.3145 and by \(0.000701\), giving: \(\Delta H^{\circ} = 0.2874 / 0.000701 \times -8.3145 \, J \, K^{-1} \, mol^{-1} = -3272.65 \, J \, mol^{-1} = -3.27 \, kJ \, mol^{-1}\). The negative sign indicates that the reaction is exothermic, releasing heat.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant (\(K\)) is crucial when studying chemical reactions that reach a state of dynamic equilibrium. Equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction.

The equilibrium constant is a number that provides the ratio of the concentration of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. For example, in a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant (\(K_c\)) is expressed as:\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] where \[\] brackets denote the concentration of a substance.

For reactions involving gases, we use the equilibrium constant in terms of partial pressure (\(K_p\)), which relates to \(K_c\) through the ideal gas law. The problem provided uses \(K_p\) since the reactants and products are gases.

Changes in temperature can shift the equilibrium position and thus change the value of the equilibrium constant. This is where the Van’t Hoff equation becomes a valuable tool—it relates the change in the equilibrium constant with temperature to the enthalpy change (\(\Delta H^\circ\)) of the reaction.

Enthalpy Change

Enthalpy change (\(\Delta H\)) is a central concept in chemical thermodynamics, representing the total heat change within a system at constant pressure. This change includes internal energy plus the product of pressure and volume.

Enthalpy is a state function, which means its change depends only on the initial and final states of the system, not on the path taken. An enthalpy change can be either positive or negative:

• If \(\Delta H > 0\), the reaction is endothermic, absorbing heat from the surroundings.
• If \(\Delta H < 0\), the reaction is exothermic, releasing heat to the surroundings.

In the given exercise, solving the Van't Hoff equation results in a negative \(\Delta H^\circ\). It tells us that the reaction \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) is exothermic.

To grasp the concept of enthalpy change more fully, it’s essential to remember that breaking bonds requires energy while forming bonds releases energy. Therefore, a reaction where bond formation exceeds bond breaking will typically release heat, indicative of an exothermic process.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationship between chemical reactions and thermodynamic laws. This field guides us in predicting whether a reaction is likely to occur spontaneously and the extent to which a reaction can proceed.

The principles of chemical thermodynamics are built upon three main laws:

• The first law of thermodynamics, also known as the law of conservation of energy, states that the total energy of an isolated system is constant; energy can be transformed from one form to another but cannot be created or destroyed.
• The second law states that for any spontaneous process, the total entropy of a system and its surroundings always increases.
• The third law suggests that as the temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.

In context to our exercise, we have used one of the tools developed from thermodynamic principles, the Van't Hoff equation, to connect the equilibrium constant and the change in temperature to the enthalpy change of the reaction. This equation provides a quantitative measure of the temperature dependence of equilibrium constants, which is an integral aspect of understanding how reactions behave under varying conditions.