Question

Hydrogen fluoride, \(\mathrm{HF}\), can be made by the reaction $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ In one experiment, \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) and \(0.050 \mathrm{~mol}\) of \(\mathrm{F}_{2}(g)\) are added to a \(0.50-\mathrm{L}\) flask. Write a reaction table for this process.

Short answer

The reaction table is set up with initial, change, and equilibrium concentrations for H\(_2\), F\(_2\), and HF as 0.20-x, 0.10-x, and 2x respectively.

Step-by-step solution

- Write the Balanced Chemical Equation

Ensure the chemical equation is balanced: equation: oindent\text{\(\mathrm{H}_{2}(g) + \mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\)}

- Calculate Initial Concentrations

Using the initial moles and the volume of the flask, calculate the initial concentrations of each reactant: oindent\text{Initial concentrations:}[H\(_2\)(g)] = \(\frac{0.10 \text{ mol}}{0.50 \text{ L}}\) = 0.20 M[F\(_2\)(g)] = \(\frac{0.050 \text{ mol}}{0.50 \text{ L}}\) = 0.10 MNo initial HF, so: [HF(g)] = 0 M

- Set Up the Reaction Table

Use a variable to represent the change in concentration for each substance: oindent\text{Let x be the change in moles for H\(_2\)(g) and F\(_2\)(g). Then, the change in moles for HF(g) will be \(2x\).}Initial: [H\(_2\)(g)] = 0.20 M, [F\(_2\)(g)] = 0.10 M, [HF(g)] = 0 MChange: [H\(_2\)(g)] = -x, [F\(_2\)(g)] = -x, [HF(g)] = +2xEquilibrium: [H\(_2\)(g)] = \(0.20 - x\), [F\(_2\)(g)] = \(0.10 - x\), [HF(g)] = \(2x\)oindent\text{Reaction Table:}\begin{array}{|c|c|c|c|}\text{Substance} & \text{Initial (M)} & \text{Change (M)} & \text{Equilibrium (M)} \hline\text{H\(_2\)(g)} & 0.20 & -x & 0.20 - x \text{F\(_2\)(g)} & 0.10 & -x & 0.10 - x \text{HF(g)} & 0.00 & +2x & 2x \end{array}

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal. This means that the concentrations of reactants and products remain constant over time. It's essential to understand that chemical equilibrium is a dynamic state, not a static one. In our given reaction

\[ \text{H}_2\text{(g)} + \text{F}_2\text{(g)} \rightleftharpoons 2 \text{HF}\text{(g)} \]

the equilibrium is achieved when the rate at which \text{H}_2 and \text{F}_2 react to form \text{HF} equals the rate at which \text{HF} decomposes back into \text{H}_2 and \text{F}_2. Thus, to analyze this balancing act, we frequently employ a reaction table. This helps to visualize the concentrations of all species at the start, during the reaction, and at equilibrium.

Initial Concentrations

Before we can understand the changes that take place during a reaction, we need to know the starting amounts of each species. These starting amounts are called the initial concentrations.
Using the information from the exercise, we have:

• Initial moles of \text{H}_2: 0.10 mol
• Initial moles of \text{F}_2: 0.050 mol
• Volume of the flask: 0.50 L

To find the initial concentrations, we use the formula:
\[ \text{Concentration} = \frac{\text{moles}}{\text{liters}} \]
Therefore, the initial concentrations are:
\[ \text{[H}_2\text{(g)]} = \frac{0.10 \text{ mol}}{0.50 \text{ L}} = 0.20 \text{ M} \] \[ \text{[F}_2\text{(g)]} = \frac{0.050 \text{ mol}}{0.50 \text{ L}} = 0.10 \text{ M} \] Since no \text{HF} is present initially, \[ \text{[HF(g)]} = 0 \text{ M} \] Calculating these initial concentrations sets the stage for determining how they change as the reaction proceeds.

Equilibrium Concentrations

Once the reaction reaches equilibrium, the concentrations of all the species will have changed from their initial values. To analyze this, we use variables to account for these changes. Let
\[ x \]
be the change in moles for \text{H}_2\text{(g)} and \text{F}_2\text{(g)}. For every mole of \text{H}_2\text{(g)} and \text{F}_2\text{(g)} that reacts, 2 moles of \text{HF} are produced.
So, the change in concentration can be written as:

• \( \text{[H}_2\text{(g)]} = 0.20 \text{ M} - x \)
• \( \text{[F}_2\text{(g)]} = 0.10 \text{ M} - x \)
• \( \text{[HF(g)]} = 2x \)

These expressions help us define the equilibrium concentrations:
\[ \text{[H}_2\text{(g)]} = 0.20 - x \] \[ \text{[F}_2\text{(g)]} = 0.10 - x \] \[ \text{[HF(g)]} = 2x \]
By setting up these relationships, we're prepared to solve for the unknowns once we have more information, such as equilibrium constants.

Balanced Chemical Equation

A balanced chemical equation is essential for correctly interpreting and solving problems in chemical equilibrium. It ensures the conservation of mass by having the same number of atoms for each element on both sides of the equation.
The given equation is
\[ \text{H}_2\text{(g)} + \text{F}_2\text{(g)} \rightleftharpoons 2 \text{HF}\text{(g)} \]
This balanced equation tells us that one mole of \text{H}_2 reacts with one mole of \text{F}_2 to produce two moles of \text{HF}. Understanding the stoichiometry (mole ratio) is crucial for setting up the reaction table and determining how the concentrations change: For every mole of \text{H}_2 and \text{F}_2 that reacts, 2 moles of \text{HF} are formed.
This mole ratio guides the changes we apply to the initial concentrations to find the equilibrium concentrations. So, always start by ensuring your chemical equation is balanced, as it impacts everything that follows.