#49 The dissociation of molecular iodine into iodine atoms is represented as1₂(g) 21 (g)-5At 1000.0 K, the equilibrium constant K for the reaction is 3.80 × 105. Suppose you start with 0.0454 mol of I₂ in a 2.25 L flask at 1000.0 K. What are theсconcentrations of the gases at equilibrium?Part 1 of 2What is the equilibrium concentration of I₂? Round your answer to 3 significant digits.Part 2 of 2Mx10MXWhat is the equilibrium concentration of I? Round your answer to 3 significant digits.x103XŚ49

equilibrium constant, Kc = 3.80 × 10-5 Initial moles of I2 = 0.0454 mol Volume = 2.25 L

The dissociation of molecular iodine into iodine atoms is represented as 1₂(g) 21 (g) At 1000.0 K, the equilibrium constant K for the reaction is 3.80 × 10. Suppose you start with 0.0454 mol of 1₂ in a 2.25 L flask at 1000.0 K. What are the concentrations of the gases at equilibrium? Part 1 of 2 What is the equilibrium concentration of I₂? Round your answer to 3 significant digits. Part 2 of 2 M X What is the equilibrium concentration of I? Round your answer to 3 significant digits. M 0 X 49

The dissociation of molecular iodine into iodine atoms is represented as 1₂(g) 21 (g) At 1000.0 K, the equilibrium constant K for the reaction is 3.80 × 10. Suppose you start with 0.0454 mol of 1₂ in a 2.25 L flask at 1000.0 K. What are the concentrations of the gases at equilibrium? Part 1 of 2 What is the equilibrium concentration of I₂? Round your answer to 3 significant digits. Part 2 of 2 M X What is the equilibrium concentration of I? Round your answer to 3 significant digits. M 0 X 49

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 12.98PAE

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