For the equations 2IBr9g) I_2(g) + Br(g) K_p = 8.5 times 10^-3 at 150 degree C I

Question

For the equations 2IBr9g) I_2(g) + Br(g) K_p = 8.5 times 10^-3 at 150 degree C If 2.8 times 10^-2 atm of IBr is placed in a 2.0L container, what is the partial pressure of IBr after equilibrium is reached? Express your answer using two significant figures and include the appropriate units? If 2.8 times 10^-7 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I_2 after equilibrium is reached? Express your answer using two significant figures and include the appropriate units?

Explanation / Answer

consider the given reaction

2IBr ---> I2 + Br2

using ICE table

at equilibrium

[pIBr] = 0.028 - 2x

[pI2] = x

[pBr2] = x

now

Kp = [pI2] [pBr2] / [pIBr]^2

so

0.0085 = [x] [x] / [ 0.028-2x]^2

0.0085 = [x]^2 / [0.028-2x]^2

0.0085 = ( [x] / [0.028-2x] )^2

( [x] / [0.028-2x] ) = 0.092195444

(0.028-2x) / x = 10.84652289

(0.028/x) - 2 = 10.84652289

0.028/x = 12.84652289

x = 2.18 x 10-3

so

at equilibrium

pI2 = x = 2.18 * 10-3

pBr2 = x = 2.18 * 10-3

pIBr = 0.028-2x = 0.028 - (2 *2.18 * 10-3) = 2.364 * 10-2 = 0.02364 atm

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