**Problem 4: Chemical Reaction Rate Analysis****A) Relate the rate of decomposition of $ N_2O_5 $ to the rate of formation of $ NO_2 $ for the following reaction in a mathematical equation.**\[ 2N_2O_5(g) \leftrightarrows 4NO_2(g) + O_2(g) \]**B) If the rate of decomposition of $ N_2O_5 $ is $ 0.484 \, \text{mol/L·s} $, what is the rate of formation of $ NO_2 $?** (5 points)---### Explanation:**Part A:**To relate the rate of decomposition of $ N_2O_5 $ to the rate of formation of $ NO_2 $, we need to use the stoichiometry of the reaction.Given the reaction:\[ 2N_2O_5(g) \leftrightarrows 4NO_2(g) + O_2(g) \]The equation indicates that 2 moles of $ N_2O_5 $ decompose to produce 4 moles of $ NO_2 $. Therefore, the rate of formation of $ NO_2 $ ($ \text{rate}_{NO_2} $) is twice the rate of decomposition of $ N_2O_5 $ ($ \text{rate}_{N_2O_5} $).Mathematically, this relationship can be described as:\[ \text{rate}_{NO_2} = 2 \times \text{rate}_{N_2O_5} \]**Part B:**We are given the rate of decomposition of $ N_2O_5 $:\[ \text{rate}_{N_2O_5} = 0.484 \, \text{mol/L·s} \]Using the relationship from Part A:\[ \text{rate}_{NO_2} = 2 \times 0.484 \, \text{mol/L·s} \]\[ \text{rate}_{NO_2} = 0.968 \, \text{mol/L·s} \]Thus, the rate of formation of $ NO_2 $ is \( 0.968 \, \text{mol/L·s

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4. A) Relate the rate of decomposition of N,0, to the rate of formation of NO, for the following reaction in a mathematical equation. 2N,O 5 - 4NO + O B) If the rate of decomposition of N,O, is 0.484 mol/liter.Sec, what is the rate of formation of NO,? (5 points)

4. A) Relate the rate of decomposition of N,0, to the rate of formation of NO, for the following reaction in a mathematical equation. 2N,O 5 - 4NO + O B) If the rate of decomposition of N,O, is 0.484 mol/liter.Sec, what is the rate of formation of NO,? (5 points)

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter11: Chemical Kinetics

Section: Chapter Questions

Problem 11.70PAE: The label on a bottle of 3% (by volume) hydrogen peroxide, H2O2, purchased at a grocery store,...

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Question

**Problem 4: Chemical Reaction Rate Analysis**

**A) Relate the rate of decomposition of $ N_2O_5 $ to the rate of formation of $ NO_2 $ for the following reaction in a mathematical equation.**

\[ 2N_2O_5(g) \leftrightarrows 4NO_2(g) + O_2(g) \]

**B) If the rate of decomposition of $ N_2O_5 $ is $ 0.484 \, \text{mol/L·s} $, what is the rate of formation of $ NO_2 $?** (5 points)

---

### Explanation:

**Part A:**

To relate the rate of decomposition of $ N_2O_5 $ to the rate of formation of $ NO_2 $, we need to use the stoichiometry of the reaction.

Given the reaction:

\[ 2N_2O_5(g) \leftrightarrows 4NO_2(g) + O_2(g) \]

The equation indicates that 2 moles of $ N_2O_5 $ decompose to produce 4 moles of $ NO_2 $.

Therefore, the rate of formation of $ NO_2 $ ($ \text{rate}_{NO_2} $) is twice the rate of decomposition of $ N_2O_5 $ ($ \text{rate}_{N_2O_5} $).

Mathematically, this relationship can be described as:

\[ \text{rate}_{NO_2} = 2 \times \text{rate}_{N_2O_5} \]

**Part B:**

We are given the rate of decomposition of $ N_2O_5 $:

\[ \text{rate}_{N_2O_5} = 0.484 \, \text{mol/L·s} \]

Using the relationship from Part A:

\[ \text{rate}_{NO_2} = 2 \times 0.484 \, \text{mol/L·s} \]

\[ \text{rate}_{NO_2} = 0.968 \, \text{mol/L·s} \]

Thus, the rate of formation of $ NO_2 $ is \( 0.968 \, \text{mol/L·s

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