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balandron [24]
3 years ago
15

If a reaction mixture initially contains 0.195 m so2cl2, what is the equilibrium concentration of cl2 at 227 ???c?

Chemistry
1 answer:
Aloiza [94]3 years ago
6 0
Missing in your question the K constant value =2.99X10^-7
The reaction equation:
         SOCl2 ⇄ SO2 +Cl2
initial 0.195          0        0
         -X             +X      +X
final  (0.195-X)      X        X

K= [SO2][Cl2]/SOCl2
by substitution
2.99X10^-7 = (X)(X)/(0.195-X)
2.99X10^-7 = X^2 / (0.195-X) by solving this equation 
∴X = 0.00024 = 2.4 X10^-4 
∴[Cl2] = 2.4X10^-4 m


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3 0
2 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
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Answer:

The new partial pressures after equilibrium is reestablished:

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PCl_5,p_3'=223.402 Torr

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PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

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