3. The decomposition of hydrogen iodide has an equilibrium constant of 1.84 x 102. What is the equilibrium pressure of hydrogen iodide if at equilibrium the hydrogen pressure and iodine pressure are 0.500 atm each? 2HIH₂ + Iz [H₂] [[2] [HI]² K= 1.84×10-² = (-5)(-5) x=3.69atm x2

Chemistry for Engineering Students
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Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.37PAE: Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was...
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3. The decomposition of hydrogen iodide has an equilibrium constant of 1.84 x 102. What is the equilibrium pressure of
hydrogen iodide if at equilibrium the hydrogen pressure and iodine pressure are 0.500 atm each? 2H = H₂+I2
[H₂] [1₂]
[HI]²
K=
1.84×10-² =
(-5)(-5)
x = 3.69atm
x²
Transcribed Image Text:3. The decomposition of hydrogen iodide has an equilibrium constant of 1.84 x 102. What is the equilibrium pressure of hydrogen iodide if at equilibrium the hydrogen pressure and iodine pressure are 0.500 atm each? 2H = H₂+I2 [H₂] [1₂] [HI]² K= 1.84×10-² = (-5)(-5) x = 3.69atm x²
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