Problem 36 - Chemistry The Central Science In Si Units

Question

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr}\). (b) Calculate \(K_{c}\).

Step-by-Step Solution

Answer

Equilibrium concentrations: \([\mathrm{H}_2] = 0.14 \ M\), \([\mathrm{Br}_2] = 0.02015 \ M\), \([\mathrm{HBr}] = 0.4 \ M\). Equilibrium constant \(K_c = 56.21\).

Step 1: Determine initial moles

Calculate the initial moles of \( \mathrm{H}_2 \) and \( \mathrm{Br}_2 \) using their molar masses. For \( \mathrm{H}_2 \), the molar mass is \( 2.02 \ g/mol \), and for \( \mathrm{Br}_2 \), the molar mass is \( 159.8 \ g/mol \). \[ n_{\mathrm{H}_2, \text{initial}} = \frac{1.374 \ g}{2.02 \ g/mol} = 0.68 \ mol \] \[ n_{\mathrm{Br}_2, \text{initial}} = \frac{70.31 \ g}{159.8 \ g/mol} = 0.4403 \ mol \]

Step 2: Determine moles at equilibrium

The moles of \( \mathrm{H}_2 \) at equilibrium are given as \( 0.566 \ g \), which is approximately \( 0.28 \ mol \). Calculate the change in moles for \( \mathrm{H}_2 \): \( \Delta n = 0.68 \ mol - 0.28 \ mol = 0.4 \ mol \). Since the reaction is 1:1:2, the change in moles for \( \mathrm{Br}_2 \) is also \( 0.4 \ mol \), and for \( \mathrm{HBr} \) it is \( 2 \times 0.4 = 0.8 \ mol \).

Step 3: Calculate equilibrium moles

Subtract the changes at equilibrium from the initial moles for \( \mathrm{Br}_2 \) and add for \( \mathrm{HBr} \). \[ n_{\mathrm{Br}_2, \text{eq}} = 0.4403 \ mol - 0.4 \ mol = 0.0403 \ mol \] \[ n_{\mathrm{HBr}, \text{eq}} = 0 + 0.8 \ mol = 0.8 \ mol \]

Step 4: Calculate equilibrium concentrations

Convert the equilibrium moles to concentrations by dividing by the volume of the vessel (2.00 L). \[ [\mathrm{H}_2] = \frac{0.28 \ mol}{2.00 \ L} = 0.14 \ M \] \[ [\mathrm{Br}_2] = \frac{0.0403 \ mol}{2.00 \ L} = 0.02015 \ M \] \[ [\mathrm{HBr}] = \frac{0.8 \ mol}{2.00 \ L} = 0.4 \ M \]

Step 5: Calculate the equilibrium constant \(K_c\)

Use the equilibrium concentrations to find \( K_c \) for the reaction: \[ K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]} = \frac{(0.4)^2}{0.14 \times 0.02015} = 56.21 \]

Key Concepts

Equilibrium Concentrations

In a chemical reaction reaching equilibrium, the concentrations of reactants and products remain constant as the rates of the forward and reverse reactions balance each other out. This steady state of concentration allows us to determine the precise amounts of substances present once equilibrium is achieved.
To find these equilibrium concentrations, you first need to know the initial amounts of the reactants and how they change as the reaction progresses. In this exercise, we begin with hydrogen (\( \mathrm{H}_2 \)) and bromine (\( \mathrm{Br}_2 \)) as the initial reactants.
The initial concentration is derived by dividing the initial number of moles by the volume of the container. As the reaction reaches equilibrium, some of these reactants are converted into the product hydrogen bromide (\( \mathrm{HBr} \)).
The equilibrium concentrations are determined through changes in the number of moles of the reactants and products. For instance, the decrease in the moles of \( \mathrm{H}_2 \) helps us calculate how many moles of \( \mathrm{Br}_2 \) and \( \mathrm{HBr} \) were formed at equilibrium. Using these values, we then derive the concentrations as moles per liter. These valuable data are essential to later calculate the equilibrium constant (\( K_c \)).

Equilibrium Constant (Kc)

The equilibrium constant \( K_c \) is a pivotal concept as it quantifies the ratio of the concentrations of products to reactants at equilibrium.
This value provides insight into the extent of a reaction, indicating how much of the reactants are converted to products. A large \( K_c \) value suggests a reaction heavily favoring products, whereas a small \( K_c \) favors reactants.
\( K_c \) is calculated using the formula:

• \( K_c = \frac{[\mathrm{Products}]^n}{[\mathrm{Reactants}]^m} \)

where \( n \) and \( m \) reflect the stoichiometric coefficients from the balanced chemical equation.
For the reaction \( \mathrm{H}_2 + \mathrm{Br}_2 \rightleftharpoons 2 \mathrm{HBr} \), the equilibrium constant equation becomes:

• \( K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]} \)

The exercise illustrates this calculation, using the concentrations derived previously to yield a \( K_c \) value of 56.21. This reflects a reaction that significantly favors the formation of \( \mathrm{HBr} \) at equilibrium. Understanding \( K_c \) clarifies reaction dynamics and potential outcomes under given conditions.

Reaction Stoichiometry

Reaction stoichiometry is the mathematical relationship between reactants and products. It's the foundation for understanding how substances interact and transform in a chemical reaction.
Stoichiometry involves balancing the chemical equation and using molar ratios to predict the quantities of substances involved. This exercise features the balanced reaction: \( \mathrm{H}_2 + \mathrm{Br}_2 \rightleftharpoons 2 \mathrm{HBr} \).
The coefficients in this equation tell us the ratio in which reactants combine and products form:

• 1 mole of \( \mathrm{H}_2 \) reacts with 1 mole of \( \mathrm{Br}_2 \) to form 2 moles of \( \mathrm{HBr} \).

This ratio helps us calculate changes in the number of moles as the reaction goes to equilibrium. For example, when \( 0.4 \) moles of \( \mathrm{H}_2 \) are consumed, \( 0.4 \) moles of \( \mathrm{Br}_2 \) are also used up, producing \( 0.8 \) moles of \( \mathrm{HBr} \).
Stoichiometry is essential for correctly determining the equilibrium amounts and concentrations, providing a clear understanding of how reactants convert to products in a balanced manner. This understanding is crucial for predicting outcomes in chemical reactions and for designing experiments in chemistry.