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 ${\mathrm{H}}_2(\mathrm{g})$ at $1.00\mathrm{atm}$ and ${\mathrm{Br}}_2(\mathrm{g})$ at $1.00\mathrm{atm}$ 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

For this reaction, the calculated equilibrium constant (K) is $2.44\times {10}^{21}$, the Gibbs free energy change (ΔG°) is $-194kJ/mol$, and the entropy change (ΔS°) is $301.9J/(mol\cdot K)$.

Step-by-step solution

Calculate the Initial Concentrations

Using the ideal gas law, we can find the initial moles of H2 and Br2 gases. We are given that both H2 and Br2 have pressures of 1.00 atm, and they are mixed in a 1.00-L flask at ${25}^{\circ }\mathrm{C}$ (or 298 K). From the ideal gas law: $PV=nRT$ We can solve for n (moles) for both H2 and Br2: $n_{H_2}=\frac{P_{H_2}V}{RT}=\frac{(1.00atm)(1.00L)}{(0.08206\frac{Latm}{molK})(298K)}=0.0408mol$ Similarly, for Br2: $n_{B{r}_2}=\frac{P_{B{r}_2}V}{RT}=0.0408mol$ Now, we can find the initial concentrations (moles per liter) of H2 and Br2: $[H_2{]}_{initial}=\frac{0.0408mol}{1L}=0.0408M$ $[B{r}_2{]}_{initial}=\frac{0.0408mol}{1L}=0.0408M$

Calculate the Equilibrium Concentrations

We are given that there are $1.10\times {10}^{13}$ molecules of H2 remaining at equilibrium. To find the equilibrium concentration of H2, we first need to find the moles of H2 remaining and then divide it by the volume of the container (1 L). Number of moles of H2 at equilibrium: $n_{H_{2(eq)}}=\frac{1.10\times {10}^{13}molecules}{6.022\times {10}^{23}\frac{molecules}{mol}}=1.83\times {10}^{-11}mol$ Equilibrium concentration of H2: $[H_2{]}_{eq}=\frac{1.83\times {10}^{-11}mol}{1L}=1.83\times {10}^{-11}M$ Since this is a 1:1 stoichiometric ratio between H2 and Br2, the change in moles of H2 must be equivalent to the change in moles of Br2. Therefore: $[B{r}_2{]}_{eq}=[B{r}_2{]}_{initial}-([H_2{]}_{initial}-[H_2{]}_{eq})=0.0408-(0.0408-1.83\times {10}^{-11})=1.83\times {10}^{-11}M$ Now, we can find the equilibrium concentration of HBr: $[HBr{]}_{eq}=2([H_2{]}_{initial}-[H_2{]}_{eq})=2(0.0408-1.83\times {10}^{-11})=0.0816M$

Calculate the Equilibrium Constant (K)

Now, using the equilibrium concentrations of all species, we can write the expression for K and calculate its value: $K=\frac{[HBr{]}^2}{[H_2][B{r}_2]}=\frac{(0.0816M{)}^2}{(1.83\times {10}^{-11}M)(1.83\times {10}^{-11}M)}=2.44\times {10}^{21}$

Calculate ΔG°

Using the calculated K value and the Gibbs-Helmholtz equation, we can calculate ΔG°: $\mathrm{\Delta }{G}^{\circ }=-RT\ln K=-(8.314\frac{J}{molK})(298K)\ln (2.44\times {10}^{21})=-193999J/mol$ Since we want the answer in kJ/mol, we can convert it: $\mathrm{\Delta }{G}^{\circ }=-194kJ/mol$

Calculate ΔS°

Finally, we can use ΔG°, ΔH°, and the temperature to calculate ΔS°: $\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$ $\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }-\mathrm{\Delta }{G}^{\circ }}{T}=\frac{-103.8kJ/mol-(-194kJ/mol)}{298K}=301.9J/(mol\cdot K)$ So, the values of K, ΔG°, and ΔS° are: $K=2.44\times {10}^{21}$ $\mathrm{\Delta }{G}^{\circ }=-194kJ/mol$ $\mathrm{\Delta }{S}^{\circ }=301.9J/(mol\cdot K)$

Key concepts

Equilibrium Constant

The equilibrium constant, often labeled as $K$, is a critical component in understanding chemical reactions and their balance at equilibrium. In essence, $K$ represents the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation at equilibrium.
For the reaction ${\mathrm{H}}_2(g)+{\mathrm{Br}}_2(g)\rightleftharpoons 2\mathrm{HBr}(g)$, the equilibrium expression is written as:

• $K=\frac{[\mathrm{HBr}{]}^2}{[{\mathrm{H}}_2][{\mathrm{Br}}_2]}$

By substituting the equilibrium concentrations of the involved species, obtained from calculations, we find that $K=2.44\times {10}^{21}$.
A large $K$ value indicates that the reaction favors product formation at equilibrium. In other words, the equilibrium position significantly lies to the right. This concept helps in predicting how much product forms in a reaction under set conditions.

Gibbs Free Energy

Gibbs Free Energy, denoted as $\mathrm{\Delta }{G}^{\circ }$, is an essential thermodynamic quantity that helps us understand the spontaneity of chemical reactions. A reaction is spontaneous under the given conditions if it has a negative $\mathrm{\Delta }{G}^{\circ }$ value.
For the reaction ${\mathrm{H}}_2(g)+{\mathrm{Br}}_2(g)\rightleftharpoons 2\mathrm{HBr}(g)$, we determined $\mathrm{\Delta }{G}^{\circ }$ using the Gibbs-Helmholtz relationship:

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

Where $R$ is the ideal gas constant (8.314 J/mol K), $T$ is the temperature in Kelvin, and $K$ is the equilibrium constant.
Substituting the known values, we find $\mathrm{\Delta }{G}^{\circ }=-194\mathrm{kJ}/\mathrm{mol}$. This negative value confirms that the reaction proceeds spontaneously towards forming hydrogen bromide (HBr) at 25°C.

Entropy Change

Entropy, represented as $\mathrm{\Delta }{S}^{\circ }$, is a measure of the disorder or randomness in a system. Understanding $\mathrm{\Delta }{S}^{\circ }$ provides insight into how the distribution of energy and matter changes during a reaction.
For the reaction ${\mathrm{H}}_2(g)+{\mathrm{Br}}_2(g)\rightleftharpoons 2\mathrm{HBr}(g)$, $\mathrm{\Delta }{S}^{\circ }$ can be calculated using the relationship:

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

Where $\mathrm{\Delta }{H}^{\circ }$ is the change in enthalpy.
For this exercise, $\mathrm{\Delta }{H}^{\circ }=-103.8\mathrm{kJ}/\mathrm{mol}$ and $\mathrm{\Delta }{G}^{\circ }=-194\mathrm{kJ}/\mathrm{mol}$, resulting in $\mathrm{\Delta }{S}^{\circ }=301.9\mathrm{J}/(\mathrm{mol}\cdot \mathrm{K})$.
This positive $\mathrm{\Delta }{S}^{\circ }$ means that the disorder increases as the reaction proceeds.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry used to relate the pressure, volume, temperature, and moles of an ideal gas. The formula is expressed as:

• $PV=nRT$
• Where $P$ is pressure, $V$ is volume, $n$ is the amount of substance in moles, $R$ is the ideal gas constant (0.08206 $L\cdot atm/(mol\cdot K)$), and $T$ is the temperature in Kelvin.

In the context of the reaction ${\mathrm{H}}_2(g)+{\mathrm{Br}}_2(g)\rightleftharpoons 2\mathrm{HBr}(g)$, the Ideal Gas Law is employed to determine the initial moles and, subsequently, the concentrations of hydrogen and bromine gases.
Given that both gases were under 1.00 atm in a 1.00 L flask at 298 K, the moles $n$ were calculated as 0.0408 mol each. These initial amounts are crucial in computing the equilibrium concentrations, which are then used to find the equilibrium constant $K$.
Thus, the Ideal Gas Law acts as a starting point for analysis in equilibrium problems.