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
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
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Chapter1: Chemical Foundations
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Could you explain how to use ICE tables and/or stoichiometry here?
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba3d1b38-49f8-49ce-9ccc-b3614f72d628%2F26c8dac1-75ac-463e-8a5f-28bf0847ca09%2Fdc1qrb1p_processed.png&w=3840&q=75)
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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