8) Consider the reactions (forward and reverse, respectively, for the same equilibrium) below: k = 2.6 x 103 liter² mol²-sec @ 380°C 2 NO(g) + O2(g) → 2 NO 2 (g) 2 NO₂ (g) → 2 NO(g) + O2(g) a. Write the equilibrium expression for the forward reaction (first reaction) and calculate the numerical value for the equilibrium constant at 380°C. Note: remember that rate= rater, and assume this is a one-step mechanism b. The system above is studied at another temperature. A 0.200 mole sample of NO2 is placed in a 5.00 liter container and allowed to come to equilibrium. When equilibrium is reached, 15.0% of the original NO2 has decomposed to NO and O2. Calculate the value for the equilibrium constant at the second temperature. liter @ 380°C kr = 4.1 mol-sec

8) Consider the reactions (forward and reverse, respectively, for the same equilibrium) below: k = 2.6 x 103 liter² mol²-sec @ 380°C 2 NO(g) + O2(g) → 2 NO 2 (g) 2 NO₂ (g) → 2 NO(g) + O2(g) a. Write the equilibrium expression for the forward reaction (first reaction) and calculate the numerical value for the equilibrium constant at 380°C. Note: remember that rate= rater, and assume this is a one-step mechanism b. The system above is studied at another temperature. A 0.200 mole sample of NO2 is placed in a 5.00 liter container and allowed to come to equilibrium. When equilibrium is reached, 15.0% of the original NO2 has decomposed to NO and O2. Calculate the value for the equilibrium constant at the second temperature. liter @ 380°C kr = 4.1 mol-sec

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 61QRT

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