Answer:
1.274 moles
Explanation:
The equation for the reaction can be represented as follows:
[tex]PCl_5(g)[/tex] ⇄ [tex]PCl_3(g)[/tex] + [tex]Cl_2(g)[/tex]
K = 0.060
K = [tex]\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
Concentration of [tex]PCl_5(g)[/tex] = [tex]\frac{numbers of moles}{volume}[/tex]
Concentration of [tex]PCl_5(g)[/tex] = [tex]\frac{3.98}{10.0}[/tex]
Concentration of [tex]PCl_5(g)[/tex] = 0.398 moles
If we construct an ICE table for the above equation; we have:
[tex]PCl_5(g)[/tex] ⇄ [tex]PCl_3(g)[/tex] + [tex]Cl_2(g)[/tex]
Initial 0.398 0 0
Change - x + x + x
Equilibrium (0.398 - x) x x
K = [tex]\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
K = [tex]\frac{[x][x]}{[0.398-x]}[/tex]
K = [tex]\frac{x^2}{0.398-x}[/tex]
0.060 = [tex]\frac{x^2}{0.398-x}[/tex]
0.06(0.398-x) = x²
0.02388 - 0.060x = x²
x² + 0.060x - 0.02388 = 0 (quadratic equation)
a = 1; b= 0.06; c= -0.02388
Using quadratic formula;
= [tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
= [tex]\frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}[/tex]
= [tex]\frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}[/tex]
= [tex]\frac{-0.06+/-\sqrt{0.09912} }{2}[/tex]
= [tex]\frac{-0.06+/-0.3148}{2}[/tex]
= [tex]\frac{-0.060+0.3148}{2}[/tex] or [tex]\frac{-0.060-0.3148}{2}[/tex]
= [tex]\frac{0.2548}{2}[/tex] or [tex]\frac{-0.3748}{2}[/tex]
= 0.1274 or -0.1874
We go by the positive value which says:
[x] = 0.1274 M
number of moles = 0.1274 × 10.0
= 1.274 moles
∴ the number of moles of Cl₂ produced at equilibrium = 1.274 moles
