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# Working with Kp for the equilibrium

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At an unspecified temperature, the equilibrium reaction system I2(g) + Cl2(g) <--> 2ICl(g) has a partial pressure of ICl(g) that is 5520 times the partial pressure of I2(g). In turn, the partial pressure of I2(g) is 2.75 times the partial pressure of Cl2(g). What is Kp for the equilibrium at this unspecified temperature?

https://brainmass.com/chemistry/chemical-equilibrium/working-kp-equilibrium-14080

## SOLUTION This solution is FREE courtesy of BrainMass!

Partial pressure of ICl = 5520* pI2

Partial Pressure of I2, pI2 = 2.75 * pCl2

therefore pICl = 5520*2.75* pCl2 = 15180 pCl2

We have, for a reaction of the form, mA+nB &lt;--&gt; xC + yD,

Kp = {(pC)^x * (pD)^y}/(pA)^m * (pB)^n

where pC is the partial pressure of C and so on.

Here Kp = (pICl)^2/(pI2 * pCl2) = (15180*pCl2)^2/(2.75 * pCl2)* (pCl2)

= 15180^2/2.75 = 83793600

Kp for the forward reaction is very high. (Please verify whether the given number is 5520 or 5.520)

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