### Problem 7.55For the reaction $ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) $, the value of $ K $ at $ 25^\circ \text{C} $ is $ 7.19 \times 10^{-3} $.#### Part ACalculate $[ \text{N}_2\text{O}_4 ]$ at equilibrium when $[ \text{NO}_2 ] = 2.40 \times 10^{-2} \, \text{mol/L}$.- Input field for $[ \text{N}_2\text{O}_4 ]$ in mol/L.- Submit button.- Request Answer link for verified response.#### Part BCalculate $[ \text{NO}_2 ]$ at equilibrium when $[ \text{N}_2\text{O}_4 ] = 7.80 \times 10^{-2} \, \text{mol/L}$.- Input field for $[ \text{NO}_2 ]$ in mol/L.- Submit button.- Request Answer link for verified response.

For the reaction N₂O4 (9) 2NO₂ (g), the value of K at 25°C is 7.19 x 10-3 ▼ Part A Calculate [N₂04] at equilibrium when [NO₂] =2.40 × 10-2mol/L 15| ΑΣΦ [N₂04] = Submit Part B Request Answer → ? Calculate [NO₂] at equilibrium when [N₂O4]-7.80 x 10-2mol/L. 195| ΑΣΦ ? mol/L

For the reaction N₂O4 (9) 2NO₂ (g), the value of K at 25°C is 7.19 x 10-3 ▼ Part A Calculate [N₂04] at equilibrium when [NO₂] =2.40 × 10-2mol/L 15| ΑΣΦ [N₂04] = Submit Part B Request Answer → ? Calculate [NO₂] at equilibrium when [N₂O4]-7.80 x 10-2mol/L. 195| ΑΣΦ ? mol/L

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

### Problem 7.55

For the reaction $ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) $, the value of $ K $ at $ 25^\circ \text{C} $ is $ 7.19 \times 10^{-3} $.

#### Part A

Calculate $[ \text{N}_2\text{O}_4 ]$ at equilibrium when $[ \text{NO}_2 ] = 2.40 \times 10^{-2} \, \text{mol/L}$.

- Input field for $[ \text{N}_2\text{O}_4 ]$ in mol/L.
- Submit button.
- Request Answer link for verified response.

#### Part B

Calculate $[ \text{NO}_2 ]$ at equilibrium when $[ \text{N}_2\text{O}_4 ] = 7.80 \times 10^{-2} \, \text{mol/L}$.

- Input field for $[ \text{NO}_2 ]$ in mol/L.
- Submit button.
- Request Answer link for verified response.

Expert Solution

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K= 7.19 x 10 ^—3

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