Respuesta :
Answer:
[Cl₂] = 0.0656 mol·L⁻¹; [PCl₅] = 0.002 37 mol·L⁻¹
Explanation:
The balanced equation is
PCl₅ ⇌ PCl₃ + Cl₂
Data:
Kc = 1.80
n = 0.197 mol
V = 2.90 L
1. Calculate the initial concentration of PCl₅
[tex]\text{[PCl$_{5}$]} = \dfrac{\text{0.197 mol}}{\text{2.90 L}} = \text{0.067 93 mol/L}\\\\[/tex]
Step 2. Set up an ICE table.
[tex]\begin{array}{lccccc} & \text{PCl}_{5} & \rightleftharpoons & \text{PCl}_{3} & + & \text{Cl}_{2} \\\text{I/mol}\cdot\text{L}^{-1}: & 0.06973 & & 0 & & 0 &\\\text{C/mol}\cdot\text{L}^{-1}: & -x & & +x & & +x &\\\text{E/mol}\cdot\text{L}^{-1}:& 0.06793-x & & x & & x &\\\end{array}[/tex]
Step 3. Calculate the equilibrium concentrations
[tex]K_{\text{c}} = \dfrac{\text{[PCl$_3$][Cl$_2$]}}{\text{[PCl$_5$]}} = \dfrac{x^{2}}{0.06793-x} = 1.80\\\\\begin{array}{rcl}\\x^{2}& = & 1.80(0.06793 - x)\\x^{2& = & 0.1223 - 1.80x\\x^{2} + 1.80x - 0.1223& = & 0\\x & = & \mathbf{0.0656}\\\end{array}[/tex]
[Cl₂] = x mol·L⁻¹ = 0.0656 mol·L⁻¹
[PCl₅] = (0.06793 - 0.0656) mol·L⁻¹ = 0.00237 mol·L⁻¹
Check:
[tex]\begin{array}{rcl}\dfrac{0.0656^{2}}{0.00237} & = & 1.80\\\\\dfrac{0.00430 }{0.00237} & = & 1.80\\\\1.81 & = & 1.80\\\end{array}[/tex]
Close enough.
The concentration of [tex]PCl_5(g)[/tex] is 0.0655 M and [tex]PCl_3(g)[/tex] is 0.00240 M at equilibrium.
Given:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g),K_c=1.80[/tex]
Moles of phosphorus pentachloride added to flask = 0.197 mol
The volume of the flask = 2.90
To find:
The concentrations of [tex]PCl_5(g)[/tex] and [tex]PCl_3(g)[/tex] at equilibrium.
Solution:
Initial concentration of phosphorus pentachloride in flask:
[tex]=\frac{0.197 mol}{2.90L}=0.0679 M[/tex]
The equilibrium constant of the reaction = [tex]K_c=1.80[/tex]
The equilibrium concentration of phosphorus trichloride = [tex][PCl_3]=x[/tex]
The equilibrium concentration of chlorine gas = [tex][Cl_2] = x[/tex]
The equilibrium concentration of phosphorus pentachloride [tex]=[PCl_5]= (0.0697- x)[/tex]
The expression of an equilibrium constant will be given as:
[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}\\\\1.80=\frac{x\times x}{(0.0679-x)}\\\\x = 0.0655[/tex]
The equilibrium concentration of phosphorus pentachloride:
[tex]=[PCl_5]= (0.0679- x) = (0.0679 -0.0655 )M=0.00240 M[/tex]
The equilibrium concentration of chlorine gas :
[tex]= [Cl_2] = x = 0.0655 M[/tex]
The concentration of [tex]PCl_5(g)[/tex] is 0.00240 M and [tex]PCl_3(g)[/tex] is 0.0655 M at equilibrium.
Learn more about the equilibrium constant here:
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