Question

Unit 5 Activity 2: Calculations Involving Keq and Factors [7 marks]
2. 1.2 mol of gaseous hydrogen fluoride (HF(g)) is injected into a 1.9 L reaction vessel and allowed to reach
equilibrium. If the equilibrium constant Keq = 7.5, calculate the equilibrium concentrations of all species
in the reaction. [7]
HF(g) ⇌ H2(g) + F2(g)

Answer 1

To find the equilibrium concentrations of all species in the reaction, we need to set up an ICE (Initial, Change, Equilibrium) table for the reaction:

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

### Step 1: Initial Concentrations
First, we calculate the initial concentration of HF.

- **Initial moles of HF = 2.1 mol**
- **Volume of the reaction vessel = 1.9 L**

The initial concentration \([HF]_0\) is calculated as follows:

\[
[HF]_0 = \frac{\text{moles}}{\text{volume}} = \frac{2.1 \, \text{mol}}{1.9 \, \text{L}} \approx 1.1053 \, \text{mol/L}
\]

The initial concentrations of H\(_2\) and F\(_2\) are both 0 since the reaction begins with only HF.

### Step 2: Set up the ICE Table
Now we can set up the ICE table:

\[
\begin{array}{|c|c|c|c|}
\hline
\text{Species} & [HF] & [H_2] & [F_2] \\
\hline
\text{Initial (mol/L)} & 1.1053 & 0 & 0 \\
\hline
\text{Change (mol/L)} & -x & +x & +x \\
\hline
\text{Equilibrium (mol/L)} & 1.1053 - x & x & x \\
\hline
\end{array}
\]

### Step 3: Express Keq
Given that the equilibrium constant \(K_{eq}\) is defined as:

\[
K_{eq} = \frac{[H_2][F_2]}{[HF]}
\]

Substituting the equilibrium concentrations from the table:

\[
7.5 = \frac{x \cdot x}{1.1053 - x} = \frac{x^2}{1.1053 - x}
\]

### Step 4: Solve the equation
Multiplying both sides by \((1.1053 - x)\):

\[
7.5(1.1053 - x) = x^2
\]
\[
8.29125 - 7.5x = x^2
\]
\[
x^2 + 7.5x - 8.29125 = 0
\]

### Step 5: Use the quadratic formula
We can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 7.5\), and \(c = -8.29125\).

\[
b^2 - 4ac = (7.5)^2 - 4(1)(-8.29125) = 56.25 + 33.165 = 89.415
\]

Now, calculate \(x\):

\[
x = \frac{-7.5 \pm \sqrt{89.415}}{2}
\]

Calculating \(\sqrt{89.415} \approx 9.45\):

\[
x = \frac{-7.5 \pm 9.45}{2}
\]

Calculating the two possible solutions for \(x\):

1. Positive solution:
\[
x = \frac{1.95}{2} \approx 0.975
\]

2. Negative solution:
\[
x = \frac{-16.95}{2} \; \text{(not physically meaningful)}
\]

We will take \(x \approx 0.975\).

### Step 6: Calculate equilibrium concentrations
Now substitute \(x\) back into the equilibrium expressions:

- \([HF] = 1.1053 - 0.975 \approx 0.1303 \, \text{mol/L}\)
- \([H_2] = x \approx 0.975 \, \text{mol/L}\)
- \([F_2] = x \approx 0.975 \, \text{mol/L}\)

### Final Answer
The equilibrium concentrations are approximately:
- \([HF] \approx 0.1303 \, \text{mol/L}\)
- \([H_2] \approx 0.975 \, \text{mol/L}\)
- \([F_2] \approx 0.975 \, \text{mol/L}\)