# chemistry questions

• Dec 31st 2008, 08:50 PM
euclid2
chemistry questions
1. what is the value of the equilibrium constant at $200^oC$ for the decomposition of phosporus pentachloride gas to phosphorus trichloride gas and chlorine gas? At equilibrium,

$[PCL_{3(g)}]=[CL_{2(g)}]=0.014 mol/L$

$[PCL_{5(g)}] = 4.3*10^{-4} mol/L$

4. In the haber process of synthesizing ammonia, the value of K is $8.00 * 10^{-7}$ for the reaction at $400^oC$. In a sealed container at equilibrium at $400^oC$ the concentations of $H_{2(g)}$ and $N_{2(g)}$ are measured to be 0.50 mol/L and 1.50mol/L respectively. Write the equilibrium constant equation and calculate the equilibreum concentration of $NH_{3(g}$)
• Dec 31st 2008, 08:57 PM
fardeen_gen
Quote:

1. what is the value of the equilibrium constant at http://www.mathhelpforum.com/math-he...4df3f347-1.gif for the decomposition of phosphate pentachloride gas to phosphorus trichloride and chlorine gas? At equilibrium,

http://www.mathhelpforum.com/math-he...fe98a82a-1.gif
The reaction is:
PCl5 <---> PCl3 + Cl2
At equilibrium: PCl5=(4.3 x 10^-4) PCl3 =0.014 Cl2= 0.014

Therefore,
Equilibrium constant K
= 0.014 x 0.014/(4.3 x 10^-4)
= 0.000196/(4.3 x 10^-4)
=0.4558
• Dec 31st 2008, 08:57 PM
euclid2
Quote:

Originally Posted by fardeen_gen
The reaction is:
PCl5 ---> PCl3 + Cl2
Initial conc. : 1 0 0
At equilibrium: 1 - x x x

Therefore, according to given data, x = 0.014 mol/L

Therefore,
Equilibrium constant K
= x.x/(1 - x)
= x^2/(1 - x)
= (0.014)^2/(1 - 0.014)
= 0.000196/0.086
= 0.0023

You were missing a bit of info, i edited, sorry.
• Dec 31st 2008, 09:14 PM
fardeen_gen
Quote:

4. In the haber process of synthesizing ammonia, the value of K is http://www.mathhelpforum.com/math-he...dac40278-1.gif for the reaction at http://www.mathhelpforum.com/math-he...fe881fba-1.gif. In a sealed container at equilibrium at http://www.mathhelpforum.com/math-he...fe881fba-1.gif the concentations of http://www.mathhelpforum.com/math-he...500abe86-1.gif and http://www.mathhelpforum.com/math-he...a20c43be-1.gif are measured to be 0.50 mol/L and 1.50mol/L respectively. Write the equilibrium constant equation and calculate the equilibreum concentration of http://www.mathhelpforum.com/math-he...322670b0-1.gif)
N2 + 3H2 <---> 2NH3
Equilibrium concentration: [H2] = 0.50 M, [N2] = 1.50 M
Let the equilibrium concentration of NH3 be x M

Therefore,
Equilibrium constant equation: (By law of mass action)
K = [NH3]^2/([H2]^3 x [N2])
8 x 10^-7 = x^2/((0.5)^3 x (1.5))
x^2 = (8 x 10^-7) x (0.125) x (1.5)
x^2 = 1.5 x 10^-7
x = 3.87 x 10^-4
• Dec 31st 2008, 09:17 PM
euclid2
Quote:

Equilibrium constant equation: (By law of mass action)
K = [NH3]^2/([H2]^3 x [N2])
x^2 = (8 x 10^-7) x (0.125) x (1.5)
these
• Dec 31st 2008, 09:19 PM
fardeen_gen
Can you quote the exact lines please?

EDIT:
By the Gulbarg-Waage law of mass action:
For a reaction:
http://upload.wikimedia.org/math/a/f...610ce90563.png where, [S], [T], [A], [B] are equlibrium concentrations and α,β, σ, τ are molar coefficients(number of moles of the particular element or compound, for example, σ moles of S are formed in the above equation)

Now, since
Quote:

the value of K is 8 x 10^-7 for the reaction at 400 degree celsius
, therefore as the required reaction also takes place at 400 degree celsius, we can use the above value of K and plug it into the equilibrium constant equation. Then all I have done is some cross multiplication(using the given values) to get concentration of NH3(which I assumed to be x moles/L)
• Dec 31st 2008, 09:21 PM
euclid2
Quote:

Equilibrium constant equation: (By law of mass action)
K = [NH3]^2/([H2]^3 x [N2])
8 x 10^-7 = x^2/((0.5)^3 x (1.5))
x^2 = (8 x 10^-7) x (0.125) x (1.5)
these
• Dec 31st 2008, 09:30 PM
euclid2
Quote:

Originally Posted by fardeen_gen
Can you quote the exact lines please?

EDIT:
By the Gulbarg-Waage law of mass action:
For a reaction: