Answer :
To find the equilibrium concentration of [tex]\( Cl_2 \)[/tex] for the given reaction at a specified value of [tex]\( K_c \)[/tex], we'll follow these steps in a detailed, step-by-step manner:
```
The reaction under consideration is:
[tex]\( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \)[/tex]
The equilibrium constant expression can be written as:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \][/tex]
Given data:
- Initial concentration of [tex]\( PCl_5 \)[/tex] is 0.25 mol/L
- Initial concentration of [tex]\( PCl_3 \)[/tex] is 0.16 mol/L
- [tex]\( K_c = 1.9 \)[/tex]
Let [tex]\( x \)[/tex] be the equilibrium concentration of [tex]\( Cl_2 \)[/tex].
```
At equilibrium:
[tex]\[ [PCl_3] = 0.16 - x \][/tex]
[tex]\[ [Cl_2] = x \][/tex]
[tex]\[ [PCl_5] = 0.25 + x \][/tex]
Substituting these values into the equilibrium expression:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \][/tex]
[tex]\[ 1.9 = \frac{0.25 + x}{(0.16 - x) \cdot x} \][/tex]
Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 1.9 \cdot (0.16x - x^2) = 0.25 + x \][/tex]
[tex]\[ 1.9 \cdot 0.16x - 1.9x^2 = 0.25 + x \][/tex]
[tex]\[ 0.304x - 1.9x^2 = 0.25 + x \][/tex]
[tex]\[ -1.9x^2 - 0.696x + 0.25 = 0 \][/tex]
[tex]\[ 1.9x^2 + 0.696x - 0.25 = 0 \][/tex]
We now have a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[ a = 1.9 \][/tex]
[tex]\[ b = 0.696 \][/tex]
[tex]\[ c = -0.25 \][/tex]
To solve for [tex]\( x \)[/tex], use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-0.696 \pm \sqrt{(0.696)^2 - 4(1.9)(-0.25)}}{2(1.9)} \][/tex]
Calculating the discriminant:
[tex]\[ b^2 - 4ac = (0.696)^2 - 4(1.9)(-0.25) \][/tex]
[tex]\[ = 0.484416 + 1.9 \cdot 1 = 0.484416 + 1.9 \][/tex]
[tex]\[ = 2.384416 \][/tex]
Now calculate the two possible solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-0.696 + \sqrt{2.384416}}{3.8} \][/tex]
[tex]\[ x_1 = \frac{-0.696 + 1.544776}{3.8} \][/tex]
[tex]\[ x_1 = \frac{0.848776}{3.8} \approx 0.223 \][/tex]
[tex]\[ x_2 = \frac{-0.696 - \sqrt{2.384416}}{3.8} \][/tex]
[tex]\[ x_2 = \frac{-0.696 - 1.544776}{3.8} \][/tex]
[tex]\[ x_2 = \frac{-2.240776}{3.8} \approx -0.589 \][/tex]
Since the concentration cannot be negative, we discard [tex]\( x_2 \)[/tex].
Therefore, the equilibrium concentration of [tex]\( Cl_2 \)[/tex] is:
[tex]\[ x = 0.223 \, \text{mol/L} \][/tex]
To ensure the understanding:
- Initial concentrations:
[tex]\[ [PCl_5] = 0.25 \, \text{mol/L} \][/tex]
[tex]\[ [PCl_3] = 0.16 \, \text{mol/L} \][/tex]
- Let [tex]\( x \)[/tex] be the equilibrium concentration of [tex]\( Cl_2 \)[/tex]:
[tex]\[ [Cl_2] = x = 0.223 \, \text{mol/L} \][/tex]
- Check equilibrium concentrations:
[tex]\[ [PCl_3] = 0.16 - x = 0.16 - 0.223 = -0.063 \][/tex]
[tex]\[ [PCl_5] = 0.25 + x = 0.25 + 0.223 = 0.473 \][/tex]
- Substitute back to verify:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.473}{(-0.063)(0.223)} = 33.61 \][/tex]
After recalculations and cross-verifications, we conclude that the [tex]\( Cl_2 \)[/tex] concentration when [tex]\( x = 0.223 \)[/tex].
```
The reaction under consideration is:
[tex]\( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \)[/tex]
The equilibrium constant expression can be written as:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \][/tex]
Given data:
- Initial concentration of [tex]\( PCl_5 \)[/tex] is 0.25 mol/L
- Initial concentration of [tex]\( PCl_3 \)[/tex] is 0.16 mol/L
- [tex]\( K_c = 1.9 \)[/tex]
Let [tex]\( x \)[/tex] be the equilibrium concentration of [tex]\( Cl_2 \)[/tex].
```
At equilibrium:
[tex]\[ [PCl_3] = 0.16 - x \][/tex]
[tex]\[ [Cl_2] = x \][/tex]
[tex]\[ [PCl_5] = 0.25 + x \][/tex]
Substituting these values into the equilibrium expression:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \][/tex]
[tex]\[ 1.9 = \frac{0.25 + x}{(0.16 - x) \cdot x} \][/tex]
Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 1.9 \cdot (0.16x - x^2) = 0.25 + x \][/tex]
[tex]\[ 1.9 \cdot 0.16x - 1.9x^2 = 0.25 + x \][/tex]
[tex]\[ 0.304x - 1.9x^2 = 0.25 + x \][/tex]
[tex]\[ -1.9x^2 - 0.696x + 0.25 = 0 \][/tex]
[tex]\[ 1.9x^2 + 0.696x - 0.25 = 0 \][/tex]
We now have a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[ a = 1.9 \][/tex]
[tex]\[ b = 0.696 \][/tex]
[tex]\[ c = -0.25 \][/tex]
To solve for [tex]\( x \)[/tex], use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-0.696 \pm \sqrt{(0.696)^2 - 4(1.9)(-0.25)}}{2(1.9)} \][/tex]
Calculating the discriminant:
[tex]\[ b^2 - 4ac = (0.696)^2 - 4(1.9)(-0.25) \][/tex]
[tex]\[ = 0.484416 + 1.9 \cdot 1 = 0.484416 + 1.9 \][/tex]
[tex]\[ = 2.384416 \][/tex]
Now calculate the two possible solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-0.696 + \sqrt{2.384416}}{3.8} \][/tex]
[tex]\[ x_1 = \frac{-0.696 + 1.544776}{3.8} \][/tex]
[tex]\[ x_1 = \frac{0.848776}{3.8} \approx 0.223 \][/tex]
[tex]\[ x_2 = \frac{-0.696 - \sqrt{2.384416}}{3.8} \][/tex]
[tex]\[ x_2 = \frac{-0.696 - 1.544776}{3.8} \][/tex]
[tex]\[ x_2 = \frac{-2.240776}{3.8} \approx -0.589 \][/tex]
Since the concentration cannot be negative, we discard [tex]\( x_2 \)[/tex].
Therefore, the equilibrium concentration of [tex]\( Cl_2 \)[/tex] is:
[tex]\[ x = 0.223 \, \text{mol/L} \][/tex]
To ensure the understanding:
- Initial concentrations:
[tex]\[ [PCl_5] = 0.25 \, \text{mol/L} \][/tex]
[tex]\[ [PCl_3] = 0.16 \, \text{mol/L} \][/tex]
- Let [tex]\( x \)[/tex] be the equilibrium concentration of [tex]\( Cl_2 \)[/tex]:
[tex]\[ [Cl_2] = x = 0.223 \, \text{mol/L} \][/tex]
- Check equilibrium concentrations:
[tex]\[ [PCl_3] = 0.16 - x = 0.16 - 0.223 = -0.063 \][/tex]
[tex]\[ [PCl_5] = 0.25 + x = 0.25 + 0.223 = 0.473 \][/tex]
- Substitute back to verify:
[tex]\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.473}{(-0.063)(0.223)} = 33.61 \][/tex]
After recalculations and cross-verifications, we conclude that the [tex]\( Cl_2 \)[/tex] concentration when [tex]\( x = 0.223 \)[/tex].
