Question

A mixture of 0.5240 M CO and 0.4380 M Cl2 is enclosed in a vessel and heated to 1000 K .

CO(g)+Cl2(g)↽−−⇀COCl2(g)c=255.0 at 1000 K

Calculate the equilibrium concentration of each gas at 1000 K .

Answer 1

Answer:

Let the initial velocity of the body be u m/s; and the constsnt acceleration be a m/s². Then displacemet (distance covered) of the body after t second is given by

s = u t + ½ a t² ———————————-(1)

The body covers 20 m in 2s and 80 m in 4 s.

20 = u×2 + ½ a 2²; => 20 = 2 u + 2 a

u + a = 10 —————————————(2)

80 = u×4 + ½ a 4²; => 80 = 4 u + 8 a; this gives,

u + 2 a = 20 ——————————————(3)

Solving (2) and (3) for u and a we get

u = 0 m/s and a = 10 m/s² ————————(4)

To obtain the distance covered in the next 4s we need to find the velocity at the beginning of this interval using,

v = u + a t => v = 0 m/s + 1.....

Answer 2

Answer:

Explanation:

                               CO             +         Cl2                   COCl2

I                            0.5420                 0.3190                     0

C                            -X                           -X                        +X

E                        0.5420 -X               0.3190 -X                  +X

Kc = [COCl2] / [CO] [Cl2]  

255   =    X / (0.5420 -X)   (0.3190 -X)

255 (0.172898 - 0.861 X + X2)  = X

255 X2 - 219.555 X + 44.08899 = X

255 X2 - 220.555 X + 44.08899 = 0

X2 - 0.8649 X  + 0.172898 = 0

X = 0.8649 + ( (0.8649)2 - 4 (1) (0.172898))1/2 / 2   ,    0.8649  - ( (0.8649)2 - 4 (1) (0.172898))1/2​​​​​​​ / 2

X = 0.8649 + [ 0.7480 - 0.6915 ] 1/2 / 2,    0.8649 - [ 0.7480 - 0.6915 ] 1/2​​​​​​​ / 2

X = (0.8649 + 0.2376) / 2, (0.8649 - 0.2376) / 2

X = 0.55, 0.313

[CO] = 0.5420 - 0.313 = 0.229 M

[Cl2] = 0.319 - 0.313 = 0.006 M

[COCl2] = 0.313 M