Calculate the number of moles of HI that are at equilibrium with 1.30 mol of H2 and 1.30 mol of I2 in a 5.50 L flask at 447°C. H2(g) + I2(g) → 2 HI(g) Kc = 66.1 at 447°C
Added by Cristian M.
Step 1
The balanced equation for the reaction is: H2(g) + I2(g) → 2 HI(g) Show more…
Show all steps
Your feedback will help us improve your experience
Dj Tan and 89 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00-L flask at 448 °C. H2 + I2 ⇌ 2HI Kc = 50.2 at 448 °C
Dj T.
The equilibrium constant, Kc, for the following system is 54.9: H2(g) + I2(g) → 2 HI(g). At equilibrium, a system contains 2.5 moles of HI and 2.12 moles of I2 in a 5 × 10–3 m3 container. Calculate the number of moles of H2(g) present at equilibrium.
John I.
A 1.000-L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448°C. The value of the equilibrium constant Kc for the reaction H2 + I2 -> 2HI at 448°C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?
Susan H.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD