In the reaction below, 4.24 atm each of H2 and I2 were placed into a 1.00...

In the reaction below, 4.24 atm each of H2 and I2 were placed into a 1.00 L flask and allowed to react: H2(g) + I2(g) <=> 2 HI(g) Given that Kp = 33.9, calculate the equilibrium pressure of I2 to 2 decimal places.

Expert Solution

H2(g) + I2(g) <=> 2 HI(g)

initial 4.24 atm 4.24 atm 0 atm

change -x -x +2x

equil 4.24-x 4.24-x 2x

Kp = pHI^2/pH2*pI2

33.9 = (2x)^2/(4.24-x)^2

x = 3.156

partial pressure of HI = 2x = 2*31.56 = 63.12 atm

partial pressure of I2 = 4.24-x

= 4.24-3.156

= 1.08 atm

partial pressure of H2 = 4.24-x

= 4.24-3.156

= 1.08 atm

queen_honey_blossom answered 2 years ago

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