At 25 ∘c, kp = 25 for the reaction h2(g)+i2(g)⇌2hi(g). at equilibrium, the partial pressure of both h2 and i2 is 0.296 atm. what is the partial pressure of hi at equilibrium?

According to this reaction:

                    H2(g)  + i2(g)      ↔   2 HI (g) 

initial          0.296      0.296           0 
 
Change         -X           -X                +2X

Equ          (0.296-X)    (0.296-X)       2 X

when Kp = p(HI)^2/p(H2)P(I2)

          25 = (2X)^2 / (0.296-X)^2

X = 0.49

∴ P(HI) =  0.49 atm 

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