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The reaction below is carried out at a different temperature at which Kc = 0.055. This time, however, the reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations of N₂, O2, and NO at equilibrium. N₂(g) + O₂(g) = 2NO(g) Express your answers to three significant figures separated by commas.

The reaction below is carried out at a different temperature at which Kc = 0.055. This time, however, the reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations of N₂, O2, and NO at equilibrium. N₂(g) + O₂(g) = 2NO(g) Express your answers to three significant figures separated by commas.

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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N2=

O2=

NO=

i got x=18.201 and that was incorrect, thanks for the help

The reaction below is carried out at a different temperature at which \( K_c = 0.055 \). This time, however, the reaction mixture starts with only the product, \([NO] = 0.0100 \, M\), and no reactants. Find the equilibrium concentrations of \( N_2 \), \( O_2 \), and \( NO \) at equilibrium. \[ \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}(g) \] **Express your answers to three significant figures separated by commas.**
Transcribed Image Text:The reaction below is carried out at a different temperature at which \( K_c = 0.055 \). This time, however, the reaction mixture starts with only the product, \([NO] = 0.0100 \, M\), and no reactants. Find the equilibrium concentrations of \( N_2 \), \( O_2 \), and \( NO \) at equilibrium. \[ \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}(g) \] **Express your answers to three significant figures separated by commas.**
Expert Solution
Step 1: Defining equilibrium constant

Answer:

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

Given data:

straight K subscript straight C equals 0.055 left square bracket NO subscript 2 right square bracket subscript initial equals 0.0100 straight M

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