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A reaction mixture at equilibrium at 175 k contains ph2 = 0.958 atm, pi2 = 0.877 atm, and phi = 0.020 atm. a second reaction mixture, also at 175 k, contains ph2 = pi2 = 0.621 atm, and phi = 0.101 atm. is the second reaction at equilibrium? if not, what will be the partial pressure of hi when the reaction reaches equilibrium at 175 k?

Respuesta :

superman1987
 from ICE table
            H2(g) +  I2 (g )↔ 2HI(g)
 equ       0.958      0.877         0.02   first mix1
              0.621        0.621         0.101      sec mix2

Kp1 = P(HI)^2 / p(H2)*p(I2) for mix 1
       = 0.02^2 / 0.958*0.877
       = 4.8x10^-4
Kp2 = P(HI)^2 / P(H2)* P(I2) for mix 2
        = 0.101^2/ 0.621*0.621 
        = 0.0265
we can see that Kp1< Kp2 that means that the sec mixture is not at equilibrium. It will go left to reduce its products and increase reactant to reduce the Kp value to achieve equilibrium.

and the partial pressure of Hi when mix 2 reach equilibrium is:
4.8x10^-4 = P(Hi)^2 / (0.621*0.621)
∴ P(Hi) at equilibrium = 0.0136 atm



okpalawalter8

The partial pressure of Hi is mathematically given as

[tex]P_(Hi) = 0.0136 atm[/tex]

What will be the partial pressure of Hi?

Generally, the equation for the Reaction   is mathematically given as

H2(g) +  I2 (g )↔ 2HI(g)

Therefore

Kp1 = P(HI)^2 / p(H2)*p(I2)

Kp1= 0.02^2 / 0.958*0.877        

Kp1= 4.8x10^-4

And

Kp2 = 0.101^2/ 0.621*0.621          

Kp2= 0.0265

Therefore, for the partial pressure of Hi

4.8x10^-4 = P(Hi)^2 / (0.621*0.621)

[tex]P_(Hi) = 0.0136 atm[/tex]

For more information on Pressure

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