Constants I Periodic Table Part A system at equilibrium contains I2(g) at a pres

Question

Constants I Periodic Table Part A system at equilibrium contains I2(g) at a pressure of 0.27 atm and I(g) at a pressure of 0.26 atm. The system is then compressed to half its volume Find the pressure of I2 when the system returns to equilibrium. Express your answer to two significant figures and include the appropriate units P,- Value Units Submit Request Answer ? Part B Find the pressure of I when the system returns to equilibrium. Express your answer to two significant figures and include the appropriate units. P 1 Value Units

Explanation / Answer

at equilibrium:

I2 = 2I
so,

Kp = (I)^2 / I2 ,

As pressure of I given=0.26 atm,

And pressure of I2 given=0.27 atm

so,

Kp = (0.26 atm)^2 / 0.27 atm = 0.2504 atm

After compressing the volume to 1/2, the pressures would double. The instantaneous partial pressure of I2 would be 0.54 atm and I would be 0.52 atm. Putting it into the equation for Kp would be 0.5004 so the system is no longer at equilibrium.

Let X be the amount that I decreases in pressure. Per the equation, the amount that the pressure of I2 increases by is X/2
So new pressure of I = 0.52 atm - X

and I2=0.54 atm - X/2,

putting that into equation of Kp gives,

0.2504 atm = (0.52 atm - X)^2 / (0.54 atm - X/2)

After solving the equation, X = 0.1848.

Pressure of I2 = 0.4476 atm
Pressure of I = 0.3352 atm
Kp = (0.26 atm)^2 / 0.27 atm = 0.2504 atm

If you suddenly compressed the volume to 1/2, the pressures would double. The instantaneous partial pressure of I2 would be 0.54 atm and I would be 0.52 atm. Putting it into the equation for Kp would give you 0.5004 so the system is no longer at equilibrium.

Let X be the amount that I decreases in pressure. Per the equation, the amount that the pressure of I2 increases by is X/2

0.2504 atm = (0.52 atm - X)^2 / (0.54 atm - X/2)

Now you just have to solve it. Clear the brackets and cross multiply, it's just ugly. I get X = 0.1848.

So new pressure of I = 0.52 atm - X

=0.52-0.1848

= 0.3352

and I2=0.54 atm - X/2,

= 0.54 atm - 0.1848/2

=0.4476
Pressure of I2 = 0.4476 atm
Pressure of I = 0.3352 atm

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