Answer:
[tex]\boxed{\text{[PCl$_{5}$] = 0.0077 mol/L; [PCl$_{3}$] = [Cl$_{2}$] = 0.117 mol/L}}[/tex]
Explanation:
The balanced equation is
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
You don't give the volume of the flask, so I assume it is 1 L.
We can set up an ICE table to organize our calculations.
[tex]\begin{array}{lccccc} & \text{PCl}_{5} & \rightleftharpoons & \text{PCl}_{3} & + & \text{Cl}_{2} \\\text{I/mol}\cdot\text{L}^{-1}: & 0.125 & & 0 & & 0 &\\\text{C/mol}\cdot\text{L}^{-1}: & -x & & +x & & +x &\\\text{E/mol}\cdot\text{L}^{-1}:& 0.125-x & & x & & x &\\\end{array}[/tex]
[tex]K_{\text{c}} = \dfrac{\text{[PCl$_3$][Cl$_2$]}}{\text{[PCl$_5$]}} = \dfrac{x^{2}}{0.125-x} = 1.80[/tex]
Check for negligibility
[tex]\dfrac{0.125}{1.80} = 0.0674 < 400.[/tex]
x is not negligible, so we must solve a quadratic.
[tex]x^{2} = 1.80(0.125 - x)\\x^{2} = 0.225 - 1.80x\\x^{2} + 1.80x - 0.225 = 0\\[/tex]
Solve for x.
x = 0.1173
[PCl₅] = 0.125 - x = 0.125 – 0.1173 = 0.0077 mol·L⁻¹
[PCl₃] = x = 0.117 mol·L⁻¹
[Cl₂] = x = 0.117 mol·L⁻¹
[tex]\boxed{\textbf{[PCl$_{5}$] = 0.0077 mol/L; [PCl$_{3}$] = [Cl$_{2}$] = 0.117 mol/L}}[/tex]
Check:
[tex]\dfrac{0.117^{2}}{0.0077} = 1.80\\\\\dfrac{0.0138}{0.0077} = 1.79[/tex]
Close enough. It checks.
In a chemical reaction when the reactants split to form two or more products are called a decomposition reaction. It is generally a breakdown of the reactants to yield products.
The concentrations of [tex]\rm PCl_{5}[/tex] is 0.0077 mol/L, [tex]\rm PCl_{3}[/tex] and [tex]\rm Cl_{2}[/tex] is 0.177 mol/L.
The balanced equation given is:
[tex]\rm PCl_{5} (g) \rightleftharpoons \rm PCl_{3} (g) + Cl_{2} (g)[/tex]
See the ICE table for the equation in the attached image below.
According to the table:
[tex]Kc = \dfrac{[\rm PCl_{3}] [Cl_{2}]}{[\rm PCl_{5}]}[/tex]
[tex]1.80 =\dfrac { x^{2} }{0.125 - x}[/tex]
Check the value for the negligibility:
[tex]\dfrac{0.125 }{1.80} = 0.0674 < 400[/tex]
From above it can be stated that it is not negligible and will be solved by the quadratic equation:
[tex]\begin{aligned}x^{2} &= 1.80 (0.125 - x)\\\\x^{2} &= 0.225 - 1.80x\\\\0 &= x^{2} + 1.80 x - 0.225\end{aligned}[/tex]
Solving further for x:
x = 0.1173
For gases:
[tex]\begin{aligned} \rm [PCl_{5}] &= 0.125 - x \\\\\rm [PCl_{5}] &= 0.125 - 0.1173 \\\\&= 0.0077 \;\rm mol L^{-1}\end{aligned}[/tex]
And,
[tex]\begin{aligned} \rm [PCl_{3}] &= 0.117 \;\rm mol L^{-1}\\\\\rm [Cl_{2}] &= 0.117 \;\rm mol L^{-1} \end{aligned}[/tex]
Therefore, the concentration of each gases are [tex]\rm PCl_{5}[/tex] is 0.0077 mol/L, [tex]\rm PCl_{3}[/tex] and [tex]\rm Cl_{2}[/tex] is 0.177 mol/L.
Learn more about the concentration of the gases here:
https://brainly.com/question/12994231
