Answered

The decomposition of PCls(g) to form PCl(g) and Clag) has a K = 33.3 at a high
temperature. If the initial concentration of PCls is 0.1500 M, what are the equilibrium
concentrations of the reactants and products?



Answer :

GinnyAnswer
Sure, let's solve this step by step.

The reaction we are dealing with is:
[tex]\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \][/tex]

Given:
- Initial concentration of PCl₅ ([tex]\([ \text{PCl}_5 ]_0\)[/tex]) = 0.1500 M
- [tex]\( K = 33.3 \)[/tex] (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) |
|------------|--------------|-------------------|------------------------|
| [tex]\(\text{PCl}_5 \)[/tex] | 0.1500 | -x | [tex]\(0.1500 - x\)[/tex] |
| [tex]\(\text{PCl}_3 \)[/tex] | 0 | +x | x |
| [tex]\(\text{Cl}_2 \)[/tex] | 0 | +x | x |

### Step 2: Write the expression for the equilibrium constant

[tex]\[ K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \][/tex]

### Step 3: Substitute the equilibrium concentrations into the equilibrium expression

[tex]\[ K = \frac{(x)(x)}{0.1500 - x} \][/tex]

Given [tex]\( K = 33.3 \)[/tex]:

[tex]\[ 33.3 = \frac{x^2}{0.1500 - x} \][/tex]

### Step 4: Solve for [tex]\( x \)[/tex]

Multiplying both sides by [tex]\((0.1500 - x)\)[/tex] to clear the fraction:

[tex]\[ 33.3 (0.1500 - x) = x^2 \][/tex]

Expanding:

[tex]\[ 33.3 \cdot 0.1500 - 33.3x = x^2 \][/tex]

Simplifying [tex]\( 33.3 \cdot 0.1500 \)[/tex]:

[tex]\[ 4.995 - 33.3x = x^2 \][/tex]

Reordering to form a standard quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]:

[tex]\[ x^2 + 33.3x - 4.995 = 0 \][/tex]

### Step 5: Solve the quadratic equation

Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:

Here, [tex]\(a = 1\)[/tex], [tex]\(b = 33.3\)[/tex], and [tex]\(c = -4.995\)[/tex]:

[tex]\[ x = \frac{-33.3 \pm \sqrt{(33.3)^2 - 4 \cdot 1 \cdot (-4.995)}}{2 \cdot 1} \][/tex]

Calculating inside the square root:

[tex]\[ x = \frac{-33.3 \pm \sqrt{1108.89 + 19.98}}{2} \][/tex]
[tex]\[ x = \frac{-33.3 \pm \sqrt{1128.87}}{2} \][/tex]
[tex]\[ x = \frac{-33.3 \pm 33.6}{2} \][/tex]

This yields two potential solutions:

[tex]\[ x = \frac{-33.3 + 33.6}{2} \approx 0.15/2 = 0.15 \][/tex]

[tex]\[ x = \frac{-33.3 - 33.6}{2} (negative value, discarded as concentrations cannot be negative) \][/tex]

So, the valid solution is:

[tex]\[ x = 0.14925 \][/tex]

### Step 6: Calculate the equilibrium concentrations

Now, substitute [tex]\( x \)[/tex] back into the expressions for equilibrium concentrations:

[tex]\[ [\text{PCl}_5]_{eq} = 0.1500 - x = 0.1500 - 0.14925 \approx 0.00075 \, \text{M} \][/tex]

[tex]\[ [\text{PCl}_3]_{eq} = x = 0.14925 \, \text{M} \][/tex]

[tex]\[ [\text{Cl}_2]_{eq} = x = 0.14925 \, \text{M} \][/tex]

### Final answer:

At equilibrium:
- [tex]\([ \text{PCl}_5 ]_{eq} \approx 0.00075 \, \text{M}\)[/tex]
- [tex]\([ \text{PCl}_3 ]_{eq} \approx 0.14925 \, \text{M}\)[/tex]
- [tex]\([ \text{Cl}_2 ]_{eq} \approx 0.14925 \, \text{M}\)[/tex]

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