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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Chrissycakes86 Chrissycakes86
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8 years ago
I have two questions i can not seem to solve can someone help me:

Phosphorus pentachloride decomposes according to the chemical equation
PCl5(g) Leftwards Arrow-> PCl3(g)+Cl2(g)          Kc=1.8 at 250 C

A 0.253 mol sample of PCl5(g) is injected into an empty 3.70L reaction vessel held at 250 C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium


The second question

At equilibrium the concentration in this system were found to be [N2]=[o2]=0.100 M and [NO}=0.500M

N2(g)+O2(g)Leftwards Arrow->2NO(g)

If more NO is added bringing its concentration to 0.800 M what will the final concentration of NO be after equilibrium is re-established


Please help thank you
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wrote...
#1
Educator
8 years ago
Quote from: Chrissycakes86 (8 years ago)
Phosphorus pentachloride decomposes according to the chemical equation
PCl5(g) -> PCl3(g)+Cl2(g)          Kc=1.8 at 250 C

A 0.253 mol sample of PCl5(g) is injected into an empty 3.70L reaction vessel held at 250 C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium

Similar Question
Phosphorus pentachloride decomposes according to the chemical equation:

PCl5(g) ====> PCl3(g) + Cl2(g) Kc=1.80 at 250 C

A 0.360 mol sample of PCl5(g) is injected into an empty 4.55 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

[PCl5]=____________M
[PCl3]=____________M

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

n(PCl₅) = 0.360 mol
M(PCl₅) = 0.360 mol / 4.55 L = 0.0791M

======> PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
initially:.... 0.0791M ⇄ 0.000M + 0.0000M
Equlbm:(0.0791 - x) ⇄ ( +x) M + (+x) M

Kc = [ (PCl₃) ] [ Cl₂ ] / [PCl₅]
(x)(x) / (0.0791 - x) = 1.80
x = 0.0759M

[PCl₅] = 0.0791 - 0.0759 = 0.0032

[PCl₅] = 0.0032M
[PCl₃] = 0.0759M
wrote...
#2
Educator
8 years ago
Quote from: Chrissycakes86 (8 years ago)
At equilibrium the concentration in this system were found to be [N2]=[o2]=0.100 M and [NO}=0.500M

N2(g)+O2(g)->2NO(g)

If more NO is added bringing its concentration to 0.800 M what will the final concentration of NO be after equilibrium is re-established
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wrote...
#3
2 years ago
THANK YOU!
Quote from: bio_man (8 years ago)
Quote from: Chrissycakes86 (8 years ago)
At equilibrium the concentration in this system were found to be [N2]=[o2]=0.100 M and [NO}=0.500M N2(g)+O2(g)->2NO(g) If more NO is added bringing its concentration to 0.800 M what will the final concentration of NO be after equilibrium is re-established
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