A 0.357 mol sample of PCl5(g) is injected into an empty 3.90 L reaction vessel h

Question

A 0.357 mol sample of PCl5(g) is injected into an empty 3.90 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Question 1 of 23 General Chemistry 4th Edition . University Science Books presented by Sapling Leaming Phosphorus pentachloride decomposes according to the chemical equation =1.80 at 250°C A 0.357 mol sample of PC15(g) is injected into an empty 3.90 L reaction vessel held at 250 Calculate the concentrations of PCs(g) nd PCI3(g) at equilibrium. Number Number PC131 = O Previous e, Check Answer (0 Next Exit | Hint

Explanation / Answer

initial concentration of PCl5 = number of mol of PCl5 / volume in L

= 0.357 mol / 3.90 L

= 0.0915 M

Let's prepare the ICE table

[PCl5] [PCl3] [Cl2]

initial 0.0915 0 0

change -1x +1x +1x

equilibrium 0.0915-1x +1x +1x

Equilibrium constant expression is

Kc = [PCl3]*[Cl2]/[PCl5]

1.8 = (1*x)(1*x)/((0.0915-1*x))

1.8 = (1*x^2)/(0.0915-1*x)

0.1647-1.8*x = 1*x^2

0.1647-1.8*x-1*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = -1

b = -1.8

c = 0.1647

solution of quadratic equation is found by below formula

x = {-b + (b^2-4*a*c)}/2a

x = {-b - (b^2-4*a*c)}/2a

b^2-4*a*c = 3.899

putting value of d, solution can be written as:

x = {1.8 + (3.899)}/-2

x = {1.8 - (3.899)}/-2

solutions are :

x = -1.887 and x = 8.727*10^-2

since x can't be negative, the possible value of x is

x = 8.727*10^-2

At equilibrium:

[PCl5] = 0.0915-x = 0.0915-0.08727 = 0.00423 M

[PCl3] = x = 0.08727 M

Answer:

[PCl5] = 0.00423 M

[PCl3] = 0.0873 M

Feel free to comment below if you have any doubts or if this answer do not work

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