Phosphorus pentachloride decomposes according to the chemical equation PCl5(g)↽−−⇀PCl3(g)+Cl2(g)????c=1.80 at 250 ∘C A 0.197 mol sample of PCl5(g) is injected into an empty 2.90 L reaction vessel held at 250 ∘C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

[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.

Tringa0 Tringa0 · Answer 2 · 2021-10-07T10:00:25+02:00

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:

brainly.com/question/9173805?referrer=searchResults

brainly.com/question/17070972?referrer=searchResults