Question

Consider the reaction $${\mathrm{H}}_2(g)+{\mathrm{Br}}_2(g)\rightleftharpoons 2\mathrm{HBr}(g)$$ where $\mathrm{\Delta }{H}^{\circ }=-103.8\mathrm{kJ}/\mathrm{mol}.$ In a particular experiment, equal moles of ${\mathrm{H}}_2(g)$ at 1.00 atm and ${\mathrm{Br}}_2(g)$ at 1.00 atm were mixed in a 1.00-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 $K,\mathrm{\Delta }{G}^{\circ },$ and $\mathrm{\Delta }{S}^{\circ }.$

Short answer

In this reaction at the given conditions, we have the equilibrium constant $K\approx 1.62\times {10}^{-9}$, $\mathrm{\Delta }{G}^{\circ }\approx -59.2\frac{\text{kJ}}{\text{mol}}$, and $\mathrm{\Delta }{S}^{\circ }\approx 149.5\frac{\text{J}}{\text{mol}\cdot \text{K}}$.

Step-by-step solution

Determine the initial moles of H2 and Br2

The initial pressures of H2 and Br2 are both 1.00 atm, which are equal. Given that the volume of the container is 1.00 L, we can use the ideal gas law ($PV=nRT$) to find the initial moles of the gases: For H2: $$n_{H_2}=\frac{P_{H_2}V}{RT}=\frac{1.00\text{atm}\cdot 1.00\text{L}}{0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\cdot (25+273.15)\text{K}}\approx 0.0409\text{mol}$$ Similarly, for Br2: $$n_{B{r}_2}=\frac{P_{B{r}_2}V}{RT}=\frac{1.00\text{atm}\cdot 1.00\text{L}}{0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\cdot (25+273.15)\text{K}}\approx 0.0409\text{mol}$$

Calculate the equilibrium concentrations of H2, Br2, and HBr

Given the number of H2 molecules at equilibrium (1.10 x 10^13), we can convert it to moles using Avogadro's number ($6.022\times {10}^{23}\text{molecules/mol}$): $$n_{H_{2eq}}=\frac{1.10\times {10}^{13}\text{molecules}}{6.022\times {10}^{23}\text{molecules/mol}}\approx 1.83\times {10}^{-11}\text{mol}$$ We can find the change in moles of H2 and Br2, and the moles of HBr formed: $$\mathrm{\Delta }{n}_{H_2}=\mathrm{\Delta }{n}_{B{r}_2}=0.0409\text{mol}-1.83\times {10}^{-11}\text{mol}\approx 0.0409\text{mol}$$ $$n_{HBr}=2\times (0.0409\text{mol}-\mathrm{\Delta }{n}_{H_2})=2\times 1.83\times {10}^{-11}\text{mol}=3.66\times {10}^{-11}\text{mol}$$ Now we can calculate the concentrations at equilibrium using the volume of the flask (1.00 L): $$[{\mathrm{H}}_2{]}_{\text{eq}}\approx \frac{1.83\times {10}^{-11}\text{mol}}{1.00\text{L}}=1.83\times {10}^{-11}\text{M}$$ $$[{\mathrm{Br}}_2{]}_{\text{eq}}\approx \frac{0.0409\text{mol}}{1.00\text{L}}=0.0409\text{M}$$ $$[\mathrm{HBr}{]}_{\text{eq}}\approx \frac{3.66\times {10}^{-11}\text{mol}}{1.00\text{L}}=3.66\times {10}^{-11}\text{M}$$

Find the equilibrium constant (K)

We can find the equilibrium constant using the equilibrium concentrations. $$K=\frac{[\mathrm{HBr}{]}_{\text{eq}}^2}{[{\mathrm{H}}_2{]}_{\text{eq}}\cdot [{\mathrm{Br}}_2{]}_{\text{eq}}}=\frac{(3.66\times {10}^{-11}\text{M}{)}^2}{(1.83\times {10}^{-11}\text{M})\cdot (0.0409\text{M})}\approx 1.62\times {10}^{-9}$$

Calculate $\mathrm{\Delta }{G}^{\circ }$ using $\mathrm{\Delta }{H}^{\circ }$ and the equilibrium constant

We can use the following relationship to find $\mathrm{\Delta }{G}^{\circ }$: $$\mathrm{\Delta }{G}^{\circ }=-RT\ln K=-8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 298.15\text{K}\ln (1.62\times {10}^{-9})\approx -59.2\frac{\text{kJ}}{\text{mol}}$$

Determine ΔS° using ΔG° and ΔH°

Finally, we can calculate the standard entropy change by rearranging the Gibbs-Helmholtz equation: $$\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }-\mathrm{\Delta }{G}^{\circ }}{T}=\frac{(-103.8\mathrm{kJ}/\mathrm{mol}-(-59.2\mathrm{kJ}/\mathrm{mol}))}{298.15\mathrm{K}}\approx 149.5\frac{\text{J}}{\text{mol}\cdot \text{K}}$$ Thus, for this reaction at given conditions, we have the equilibrium constant $K\approx 1.62\times {10}^{-9}$, $\mathrm{\Delta }{G}^{\circ }\approx -59.2\frac{\text{kJ}}{\text{mol}}$, and $\mathrm{\Delta }{S}^{\circ }\approx 149.5\frac{\text{J}}{\text{mol}\cdot \text{K}}$.

Key concepts

Equilibrium Constant (K)

In chemical reactions, reaching equilibrium is like finding a balance. For any reversible reaction, the equilibrium constant, symbolized as $K$, tells us the ratio of product concentrations to reactant concentrations when the reaction is at equilibrium. This constant is specific to a particular reaction at a given temperature.
Typically, for a reaction of the type $aA+bB\rightleftharpoons cC+dD$, the equilibrium constant $K$ is defined by the expression:

• $K=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

"Products over reactants" is the basic rule for calculating $K$. It's important because it helps us predict the direction the reaction will shift if disturbed. When $K$ is large, products are favored. When $K$ is small, like in our problem ($1.62\times {10}^{-9}$), it indicates that reactants are favored.
Understanding how to calculate $K$ is key to predicting the concentrations of substances at equilibrium, providing insight into the reaction's behavior.

Gibbs Free Energy (ΔG°)

Gibbs Free Energy, represented as $\mathrm{\Delta }{G}^{\circ }$, is a thermodynamic quantity that helps us understand whether a reaction is spontaneous. A negative $\mathrm{\Delta }{G}^{\circ }$ means a reaction can occur without needing additional energy — it's spontaneous. Conversely, a positive $\mathrm{\Delta }{G}^{\circ }$ means the reaction requires energy input.
The relationship between $\mathrm{\Delta }{G}^{\circ }$ and the equilibrium constant $K$ is given by the expression:

• $\mathrm{\Delta }{G}^{\circ }=-RT\ln K$

Here, $R$ is the universal gas constant and $T$ is the temperature in Kelvin. In our solved exercise, $\mathrm{\Delta }{G}^{\circ }$ was calculated to be $-59.2\mathrm{kJ}/\mathrm{mol}$, showing that the reaction is spontaneous under the given conditions.
This connection between $\mathrm{\Delta }{G}^{\circ }$ and $K$ allows chemists to determine the tendency of reactions to occur and adjust reaction conditions accordingly to make desired products.

Entropy Change (ΔS°)

Entropy Change, $\mathrm{\Delta }{S}^{\circ }$, tells us about the disorder or randomness in a system. When a reaction results in more disorder (positive $\mathrm{\Delta }{S}^{\circ }$), the system becomes more favorable. If it decreases disorder (negative $\mathrm{\Delta }{S}^{\circ }$), the system tends to be less favorable.
The change in standard entropy can be determined by using the equation:

• $\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }-\mathrm{\Delta }{G}^{\circ }}{T}$

For the reaction provided in the exercise, $\mathrm{\Delta }{S}^{\circ }$ was found to be $149.5\mathrm{J}/\mathrm{mol}\cdot \mathrm{K}$. This positive value indicates an increase in the disorder during the reaction, which aligns with the spontaneous nature of the process described by $\mathrm{\Delta }{G}^{\circ }$.
Understanding entropy is crucial for predicting how changes in pressure, volume, temperature, or concentration impact the spontaneity and progression of chemical reactions. This concept helps chemists devise strategies for optimizing reaction conditions.