Question

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{B} \mathrm{r}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol}\) . In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 \(\mathrm{atm}\) and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a \(1.00-\mathrm{L}\) flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

Short answer

For the given reaction, we initially have: Moles of H2 = \((1.10 * 10^{13}) / (6.022 * 10^{23})\) Moles of Br2 = 1 - Moles of H2 Moles of HBr = 2 * Moles of H2 Now calculate the concentrations and the equilibrium constant (K): K = \(\frac{[HBr]^2}{[H2][Br2]}\) Then calculate ΔG° using: ΔG° = -RT * ln(K) Lastly, evaluate ΔS° with the following equation: ΔS° = \(\frac{ΔH° - ΔG°}{T}\) After performing these calculations, you will obtain the values of K, ΔG°, and ΔS° for the given reaction.

Step-by-step solution

Determine the number of moles of each species at equilibrium

Analyze the data from the experiment and find out the number of moles of each species at equilibrium. We know that there are 1.10x10^13 molecules of H2; we can use Avogadro's number to find the number of moles: Moles of H2 = (1.10 * 10^13) / (6.022 x 10^23) Now, we can use the stoichiometry of the reaction to find the balanced equation and calculate the moles of Br2 and HBr at equilibrium as well: 1. Moles of Br2 = Moles_initial - Moles_reaction = Moles_initial - Moles_H2_equilibrium = 1 - Moles of H2 2. Moles of HBr = Moles_initial + Moles_reaction = Moles_initial + 2 * Moles_H2_equilibrium = 2 * Moles of H2 We can also find their concentrations by dividing by the flask's volume (1.00 L), which is the same for all of them.

Calculate the equilibrium constant (K)

Using the concentrations of the species at equilibrium, we can calculate the equilibrium constant (K). The balanced equation is: H2(g) + Br2(g) ⇌ 2 HBr(g) So, K = [HBr]^2 / ([H2] * [Br2]) Plug in the concentrations you calculated in step 1 to find K.

Calculate ΔG°

Now that we have K, we can use ΔG° = -RT * ln(K) to calculate the standard Gibbs free energy change. The temperature is given in the problem as 25°C, which is 298 K in Kelvin. The gas constant R = 8.314 J/mol K. Plug in these values to get ΔG°.

Calculate ΔS°

Finally, we can use the relationship between ΔG°, ΔH°, and ΔS° to determine ΔS°. The standard enthalpy change ΔH° is given in the problem as -103.8 kJ/mol. To find ΔS°, use: ΔG° = ΔH° - T * ΔS° Rearrange the formula for ΔS°: ΔS° = (ΔH° - ΔG°) / T Plug in the values of ΔH°, ΔG°, and T to find ΔS°. Make sure to convert ΔH° to J/mol before dividing. This will give you the values of K, ΔG°, and ΔS° for the reaction.

Key concepts

Gibbs Free Energy

In chemical reactions, Gibbs Free Energy (\( \Delta G^{\circ} \)) plays a critical role in determining whether a reaction is spontaneous. At its core, Gibbs Free Energy is a thermodynamic potential that combes enthalpy and entropy. It determines the amount of work that can be performed by a thermodynamic process.
The equation to calculate the change in Gibbs Free Energy is: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\] where:

• \( \Delta H^{\circ} \) is the change in enthalpy.
• \( T \) is the temperature in Kelvin.
• \( \Delta S^{\circ} \) is the change in entropy.

Gibbs Free Energy determines if a process is spontaneous at constant temperature and pressure. If \( \Delta G^{\circ} \) is negative, the process proceeds spontaneously. For the reaction considered, calculating \( \Delta G^{\circ} \) is crucial to understand its spontaneity under specified conditions.Furthermore, the relationship with the equilibrium constant (\( K \)) is described by:\[\Delta G^{\circ} = -RT \ln(K)\]This formula links the spontaneity of the reaction to its equilibrium state, highlighting the deep connection between thermodynamics and reaction kinetics.

Reaction Stoichiometry

Reaction stoichiometry is fundamental when calculating equilibrium conditions. It refers to the quantitative relationship between reactants and products in a chemical reaction. Stoichiometry allows us to predict the amounts of substances consumed and produced.
For example, the balanced reaction \(\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g)\) provides a key mapping: one mole of \(\text{H}_2\) reacts with one mole of \(\text{Br}_2\) to produce two moles of \(\text{HBr}\).To find the equilibrium concentrations, you start by determining the initial moles and how they change as the reaction progresses. At equilibrium, a small amount of reactants might remain. Using stoichiometry, you can relate the change in this amount to the formation of products. The initial and equilibrium amounts help in determining the equilibrium constant (\( K \)) by using concentration ratios. Evaluating the equilibrium concentrations provides insight into the reaction's progress and extent under given conditions.

Enthalpy and Entropy Changes

Understanding enthalpy (\( \Delta H^{\circ} \)) and entropy (\( \Delta S^{\circ} \)) changes is key to predicting the feasibility and behavior of chemical reactions. Enthalpy change measures heat absorbed or released under constant pressure. Negative enthalpy suggests an exothermic reaction, releasing energy. In contrast, entropy change measures the disorder or randomness in a system. An increase in entropy (\( \Delta S^{\circ} > 0\)) indicates greater disorder. Reactions may favor higher entropy as systems tend to move toward higher randomness.Both enthalpy and entropy directly affect Gibbs Free Energy (\( \Delta G^{\circ} \)), thereby influencing reaction spontaneity:

• If \( \Delta H^{\circ} \) is negative and \( \Delta S^{\circ} \) is positive, \( \Delta G^{\circ} \) is more likely negative, promoting spontaneity.
• These variables can also counteract. A dominant positive \( \Delta S^{\circ} \) can drive a reaction forward even with a positive \( \Delta H^{\circ} \).

By analyzing both concepts, chemists gain better insights into reaction pathways and conditions favorable for industrial and everyday chemical processes.