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 \(x\):
-
Positive solution: \[ x = \frac{1.95}{2} \approx 0.975 \]
-
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:
- \([HF] \approx 0.1303 , \text{mol/L}\)
- \([H_2] \approx 0.975 , \text{mol/L}\)
- \([F_2] \approx 0.975 , \text{mol/L}\)