Question

Consider the following isotope exchange reaction: \\[ \mathrm{DCl}(g)+\operatorname{HBr}(g) \rightleftharpoons \operatorname{DBr}(g)+\mathrm{HCl}(g) \\] The amount of each species at equilibrium can be measured using proton and deuterium NMR (see Journal of Chemical Education \(73[1996]: 99) .\) Using the following spectroscopic information, determine \(K_{p}\) for this reaction at \(298 \mathrm{K}\). For this reaction, \(\Delta \varepsilon=41 \mathrm{cm}^{-1}\) equal to the difference in zero-point energies between products versus reactants, and the groundstate electronic degeneracy is zero for all species. $$\begin{array}{lccc} & M\left(\mathrm{g} \mathrm{mol}^{-1}\right) & B\left(\mathrm{cm}^{-1}\right) & \widetilde{\nu}\left(\mathrm{cm}^{-1}\right) \\ \hline \mathrm{H}^{35} \mathrm{Cl} & 35.98 & 10.59 & 2991 \\ \mathrm{D}^{35} \mathrm{Cl} & 36.98 & 5.447 & 2145 \\ \mathrm{H}^{81} \mathrm{Br} & 81.92 & 8.465 & 2649 \\ \mathrm{D}^{81} \mathrm{Br} & 82.93 & 4.246 & 1885 \end{array}$$

Short answer

The equilibrium constant, \(K_p\), for the given isotope exchange reaction at 298 K is approximately \(8.95 \times 10^{-8}\).

Step-by-step solution

1. Write the equilibrium constant expression (Kp)

The equilibrium constant expression for the given reaction is: \[ K_p = \frac{[DBr][HCl]}{[DCl][HBr]} \] where the brackets denote the equilibrium molar concentrations of each species.

2. Calculate the Gibbs free energy change of the reaction at 298K using the given zero-point energy difference (∆ε)

We are given the zero-point energy difference (∆ε) between the products and reactants as 41 cm⁻¹. We need to convert this into Gibbs free energy change (∆G) using the relation: \[ \Delta G = \Delta \varepsilon \cdot h \cdot c \] Where: \(\Delta G\) = Gibbs free energy change in J/mol \(\Delta \varepsilon\) = Zero-point energy difference in cm⁻¹ h = Planck's constant in J.s = 6.626 x 10⁻³⁴ J.s c = Speed of light in cm/s = 3 x 10¹⁰ cm/s Calculate ∆G at 298K: \[ \Delta G = 41 \, \text{cm}^{-1} \times (6.626 \times 10^{-34} \, \text{J s})(3 \times 10^{10} \, \text{cm s}^{-1}) = 8.17 \times 10^{-21} \, \text{J} \] \[ \Delta G = (8.17 \times 10^{-21} \, \text{J})\, \times (6.022 \times 10^{23} \, \text{mol}^{-1}) = 4916 \, \text{J mol}^{-1} \]

3. Calculate the standard Gibbs free energy change (∆G°) of the reaction using the equilibrium constant

We can calculate the standard Gibbs free energy change (∆G°) of the reaction at 298K using the relation: \[ \Delta G^{\circ} = -RT \ln(K_p) \] where: \(\Delta G^{\circ}\) = Standard Gibbs free energy change in J/mol R = Gas constant in J/(mol K) = 8.314 J/(mol K) T = Temperature in K = 298 K \(K_p\) = Equilibrium constant Rearrange the equation to solve for \(K_p\): \[ K_p = \exp\left(-\frac{\Delta G^{\circ}}{RT}\right) \] Now, substitute the values of ∆G°, R, and T: \[ K_p = \exp\left(-\frac{4916 \, \text{J mol}^{-1}}{(8.314 \, \text{J mol}^{-1} \text{K}^{-1})(298 \, \text{K})}\right) \] \[ K_p \approx 8.95 \times 10^{-8} \] So the equilibrium constant, \(K_p\), for the isotope exchange reaction at 298 K is approximately \(8.95 \times 10^{-8}\).

Key concepts

Isotope Exchange Reaction

An isotope exchange reaction is a chemical process in which isotopes of an element are exchanged between compounds. This type of reaction is particularly useful in studying reaction mechanisms and the rates of chemical reactions. It involves isotopes because they behave almost identically in chemical reactions, with only slight differences due to their mass differences.

Consider the given reaction:
\[ \mathrm{DCl}(g) + \operatorname{HBr}(g) \rightleftharpoons \operatorname{DBr}(g) + \mathrm{HCl}(g) \]
Here, deuterium (\mathrm{D}), a stable isotope of hydrogen, is exchanged between hydrogen chloride (HCl) and hydrogen bromide (HBr). The purity and specificity of isotopic labeling allow researchers to trace the path of atoms through a reaction, providing insights into the reaction dynamics.

When dealing with isotope exchange reactions, it is essential to note that the physical properties of isotopes can affect reaction rates and equilibrium positions. For instance, the heavier isotopic species, in general, will have lower zero-point energies, which relates to the concept of the Gibbs free energy in a reaction.

NMR Spectroscopy

NMR spectroscopy, or Nuclear Magnetic Resonance spectroscopy, is a powerful analytical technique used to determine the structure, dynamics, reaction state, and chemical environment of molecules. In the context of isotope exchange reactions, NMR can quantitatively analyze the concentration of reactants and products at equilibrium.

For instance, in our exercise, proton and deuterium NMR are used to measure the amount of each species at equilibrium, which is crucial for determining the equilibrium constant \( K_p \). The spectroscopic information provided includes parameters such as molecular weight (\mathrm{M}), rotational constants (\mathrm{B}), and wavenumbers (\widetilde{u}) that are characteristic of the specific compounds involved. These parameters are a direct result of molecular structure and thus are indicative of the unique environment of the nuclei within the molecule.

NMR spectroscopy's ability to distinguish between isotopes based on their magnetic properties is what makes it so invaluable in studying reactions like isotope exchange, giving precise data that can feed into calculations of chemical equilibrium.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a thermodynamic quantity that is a measure of the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. It is a vital concept as it determines the spontaneity of a reaction; reactions tend toward lower Gibbs free energy.

In our exercise, we calculate the Gibbs free energy change of the reaction using the given zero-point energy difference (\( \Delta \varepsilon \)) and its relation to \( \Delta G \). This difference is tied to the intrinsic vibrational energy levels of the isotopic molecules – the 'zero-point' being the lowest vibrational state they can achieve. The relation between \( \Delta \varepsilon \) and \( \Delta G \) involves Planck’s constant (\( h \)) and the speed of light (\( c \)), grounding the equation in fundamental physical constants.

Furthermore, understanding the relationship between Gibbs free energy and the equilibrium constant (\( K_p \)) is critical for quantifying the extent of the reaction. The negative natural logarithm of the equilibrium constant, multiplied by the gas constant (\( R \)) and the temperature (\( T \)), gives the standard Gibbs free energy (\( \Delta G^\circ \)). This allows chemists to predict the direction of a chemical reaction and determine the conditions under which the system reaches equilibrium based on the sign and magnitude of \( \Delta G \).