Sure, let's solve this step by step.
The reaction we are dealing with is:
\(\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]\)
Given:
- Initial concentration of PCl₅ (\(\([ \text{PCl}_5 ]_0\)\)) = 0.1500 M
- \(\( K = 33.3 \)\) (equilibrium constant)
We want to find the equilibrium concentrations of PCl₅, PCl₃, and Cl₂.
### Step 1: Set up the ICE Table
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|------------|--------------|-------------------|------------------------|
| \(\(\text{PCl}_5 \)\) | 0.1500 | -x | \(\(0.1500 - x\)\) |
| \(\(\text{PCl}_3 \)\) | 0 | +x | x |
| \(\(\text{Cl}_2 \)\) | 0 | +x | x |
### Step 2: Write the expression for the equilibrium constant
\(\[ K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]\)
### Step 3: Substitute the equilibrium concentrations into the equilibrium expression
\(\[ K = \frac{(x)(x)}{0.1500 - x} \]\)
Given \(\( K = 33.3 \)\):
\(\[ 33.3 = \frac{x^2}{0.1500 - x} \]\)
### Step 4: Solve for \(\( x \)\)
Multiplying both sides by \(\((0.1500 - x)\)\) to clear the fraction:
\(\[ 33.3 (0.1500 - x) = x^2 \]\)
Expanding:
\(\[ 33.3 \cdot 0.1500 - 33.3x = x^2 \]\)
Simplifying \(\( 33.3 \cdot 0.1500 \)\):
\(\[ 4.995 - 33.3x = x^2 \]\)
Reordering to form a standard quadratic equation \(\(ax^2 + bx + c = 0\)\):
\(\[ x^2 + 33.3x - 4.995 = 0 \]\)
### Step 5: Solve the quadratic equation
Using the quadratic formula \(\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)\):
Here, \(\(a = 1\)\), \(\(b = 33.3\)\), and \(\(c = -4.995\)\):
\(\[ x = \frac{-33.3 \pm \sqrt{(33.3)^2 - 4 \cdot 1 \cdot (-4.995)}}{2 \cdot 1} \]\)
Calculating inside the square root:
\(\[ x = \frac{-33.3 \pm \sqrt{1108.89 + 19.98}}{2} \]\)
\(\[ x = \frac{-33.3 \pm \sqrt{1128.87}}{2} \]\)
\(\[ x = \frac{-33.3 \pm 33.6}{2} \]\)
This yields two potential solutions:
\(\[ x = \frac{-33.3 + 33.6}{2} \approx 0.15/2 = 0.15 \]\)
\(\[ x = \frac{-33.3 - 33.6}{2} (negative value, discarded as concentrations cannot be negative) \]\)
So, the valid solution is:
\(\[ x = 0.14925 \]\)
### Step 6: Calculate the equilibrium concentrations
Now, substitute \(\( x \)\) back into the expressions for equilibrium concentrations:
\(\[ [\text{PCl}_5]_{eq} = 0.1500 - x = 0.1500 - 0.14925 \approx 0.00075 \, \text{M} \]\)
\(\[ [\text{PCl}_3]_{eq} = x = 0.14925 \, \text{M} \]\)
\(\[ [\text{Cl}_2]_{eq} = x = 0.14925 \, \text{M} \]\)
### Final answer:
At equilibrium:
- \(\([ \text{PCl}_5 ]_{eq} \approx 0.00075 \, \text{M}\)\)
- \(\([ \text{PCl}_3 ]_{eq} \approx 0.14925 \, \text{M}\)\)
- \(\([ \text{Cl}_2 ]_{eq} \approx 0.14925 \, \text{M}\)\)
