**Equilibrium Concentration Calculation for NO**When 2.0 M NO, N₂, and O₂ are added to a reaction container of 1.0 L at 2275 K, they react according to the balanced chemical equation:\[ 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \]Given:- Equilibrium constant (K) = 2400- Fraction of NO initially dissociated (X) = 0.97The task is to determine the equilibrium concentration of NO. Given the equilibrium expression for this reaction:\[ K = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} \]To find \([ \text{NO}]\) at equilibrium, we use the initial concentrations, the change in concentration, and the equilibrium constant.**Step-by-Step Calculation:**1. Let the initial concentration of NO be \( 2.0 \text{ M} \).2. Let \(X\) be the fraction that NO dissociates, which is 0.97, so \( 2.0 \times 0.97 \) will dissociate.3. The initial concentrations of N₂ and O₂ are both 0 M.4. Because the dissociation ratio is 1:1, and for every 2 moles of NO that dissociate, 1 mole of N₂ and 1 mole of O₂ are formed,5. At equilibrium, the concentration of NO will be reduced by \( 2.0 \times 0.97 / 2 = 1.97 \text{ M} \).Calculate the equilibrium concentrations for:\[ \text{[NO]} = 2.0 - 2(0.97) = 0.06 \text{ M}\]\[ \text{[N}_2\text{]} = 0 + 0.97 = 0.97 \text{ M}\]\[ \text{[O}_2\text{]} = 0 + 0.97 = 0.97 \text{ M}\]Finally, plug the equilibrium concentrations into the equilibrium expression.\[ "K" = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]

In ICE Table, I represent initial concentration C is change in concentration E is concentration at…

2.0 M NO, N2, and O2 are added to a 1.0 L container at 2275 K and react according to the following. 2NO(g) = N₂(g) + O₂(g) K = 2400 "X" = 0.97 What is the concentration of NO at equilibrium? [NO] = [?] M [NO], M Enter

2.0 M NO, N2, and O2 are added to a 1.0 L container at 2275 K and react according to the following. 2NO(g) = N₂(g) + O₂(g) K = 2400 "X" = 0.97 What is the concentration of NO at equilibrium? [NO] = [?] M [NO], M Enter

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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Question

**Equilibrium Concentration Calculation for NO**

When 2.0 M NO, N₂, and O₂ are added to a reaction container of 1.0 L at 2275 K, they react according to the balanced chemical equation:

\[ 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \]

Given:
- Equilibrium constant (K) = 2400
- Fraction of NO initially dissociated (X) = 0.97

The task is to determine the equilibrium concentration of NO. Given the equilibrium expression for this reaction:

\[ K = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} \]

To find \([ \text{NO}]\) at equilibrium, we use the initial concentrations, the change in concentration, and the equilibrium constant.

**Step-by-Step Calculation:**
1. Let the initial concentration of NO be \( 2.0 \text{ M} \).
2. Let \(X\) be the fraction that NO dissociates, which is 0.97, so \( 2.0 \times 0.97 \) will dissociate.
3. The initial concentrations of N₂ and O₂ are both 0 M.
4. Because the dissociation ratio is 1:1, and for every 2 moles of NO that dissociate, 1 mole of N₂ and 1 mole of O₂ are formed,
5. At equilibrium, the concentration of NO will be reduced by \( 2.0 \times 0.97 / 2 = 1.97 \text{ M} \).

Calculate the equilibrium concentrations for:
\[ \text{[NO]} = 2.0 - 2(0.97) = 0.06 \text{ M}\]
\[ \text{[N}_2\text{]} = 0 + 0.97 = 0.97 \text{ M}\]
\[ \text{[O}_2\text{]} = 0 + 0.97 = 0.97 \text{ M}\]

Finally, plug the equilibrium concentrations into the equilibrium expression.
\[ "K" = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]

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