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A student ran the following reaction in the laboratory at 663 K: 2 NH₂(g) → N₂(g) + 3 H₂(g) When she introduced 6.71E-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 4.34E-3 M. Calculate the equilibrium constant, K., she obtained for this reaction. K₂=

A student ran the following reaction in the laboratory at 663 K: 2 NH₂(g) → N₂(g) + 3 H₂(g) When she introduced 6.71E-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 4.34E-3 M. Calculate the equilibrium constant, K., she obtained for this reaction. K₂=

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A student ran the following reaction in the laboratory at 663 K: 2 NH3(g) = N₂(g) + 3 H₂(g) When she introduced 6.71E-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 4.34E-3 M. Calculate the equilibrium constant, K., she obtained for this reaction. K₂ =
Transcribed Image Text:A student ran the following reaction in the laboratory at 663 K: 2 NH3(g) = N₂(g) + 3 H₂(g) When she introduced 6.71E-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 4.34E-3 M. Calculate the equilibrium constant, K., she obtained for this reaction. K₂ =
Expert Solution
Step 1: Defining equilibrium constant

Answer:

For any reaction, value of equilibrium constant is equal to the ratio of molar concentrations of products and molar concentration of reactants.

Given data:

straight n subscript NH subscript 3 left parenthesis initial right parenthesis end subscript equals 6.71 cross times 10 to the power of negative 2 end exponent mol left square bracket NH subscript 3 right square bracket subscript equilibrium equals 4.34 cross times 10 to the power of negative 3 end exponent straight M

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