Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 141

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{kJ}\) . If $2.50 \times 10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\( at \)100 .^{\circ} \mathrm{C}$ and 3.50 atm and \(4.50 \times\) \(10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at $100 .^{\circ} \mathrm{C}$ and 3.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

Short Answer

Expert verified
The heat required to synthesize the maximum yield of NO₂ when given volumes and pressures of N₂ and O₂ is 1.938 kJ.

Step by step solution

01

1. Balance the Chemical Equation

To balance the equation, start by counting the number of atoms of each element on the reactant and product sides. Based on the given reaction, we have: N₂(g) + O₂(g) → NO₂(g) (unbalanced) Adjust the coefficients to balance the number of atoms on each side of the equation: 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) The balanced coefficient is 2:2:4 for N₂, O₂ and NO₂, respectively.
02

2. Convert mL Gas to Moles Using Ideal Gas Law

Next, we will convert the given volumes and pressures of N₂ and O₂ into moles using the Ideal Gas Law, PV=nRT. For N₂: - Volume (V) = 2.50 x 10² mL = 0.25 L - Pressure (P) = 3.50 atm - Temperature (T) = 100 °C = 373.15 K - Gas constant (R) = 0.0821 L atm / (K mol) - n (moles N₂) = ? PV = nRT --> n = PV / (RT) n(N₂) = (3.50 atm * 0.25 L) / (0.0821 L atm / (K mol) * 373.15 K) n(N₂) = 0.0286 mol For O₂: - Volume (V) = 4.50 x 10² mL = 0.45 L - Pressure (P) = 3.50 atm - Temperature (T) = 100 °C = 373.15 K - Gas constant (R) = 0.0821 L atm / (K mol) - n (moles O₂) = ? PV = nRT --> n = PV / (RT) n(O₂) = (3.50 atm * 0.45 L) / (0.0821 L atm / (K mol) * 373.15 K) n(O₂) = 0.0514 mol
03

3. Determine the Limiting Reactant

Determine the limiting reactant by comparing the stoichiometric coefficients of N₂ and O₂ according to the balanced chemical equation: 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) n(N₂) = 0.0286 mol n(O₂) = 0.0514 mol Ratio of moles of N₂ to O₂: (0.0286 mol N₂/0.0514 mol O₂) = 0.556 Since the ratio of moles N₂:O₂ is less than the balanced equation's stoichiometric coefficient (2:2), N₂ is the limiting reactant.
04

4. Calculate the Maximum Yield of NO₂

The maximum yield of NO₂ can be found using the stoichiometric ratio from the balanced chemical equation. 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) Using the limiting reactant N2 (0.0286 mol), calculate the moles of NO₂: n(NO₂) = 0.0286 mol N₂ * (4 mol NO₂ / 2 mol N₂) n(NO₂) = 0.0572 mol NO₂
05

5. Calculate the Heat Required

Finally, calculate the heat required for the reaction, using the enthalpy change of reaction, ΔH, provided: ΔH = 67.7 kJ (1 mol N₂ reacts with 1 mol O₂) Since there are 2 moles of N₂ as coefficient, we have: Heat required for 2 mole of N₂ = 67.7 kJ\(\times2\)=135.4 kJ Calculate the heat required for 0.0286 moles of N₂ (limiting reactant) by proportion: Heat Required = (0.0286 mol N₂ * 135.4 kJ) / 2 mol N₂ Heat Required = 1.938 kJ Thus, 1.938 kJ of heat is required to synthesize the maximum yield of NO₂.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each of the following situations a-c, use choices i-iii to complete the statement: "The final temperature of the water should be.." i. between \(50^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) . ii. \(50^{\circ} \mathrm{C}\) . iii. between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) . a. 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a 100.0 -g sample of water at \(10^{\circ} \mathrm{C}\) . b. A 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a $500.0 . \mathrm{g}\( sample of water at \)10^{\circ} \mathrm{C} .$ c. You have a Styrofoam cup with 50.0 \(\mathrm{g}\) of water at $10^{\circ} \mathrm{C}\( . You add a 50.0 -g iron ball at \)90^{\circ} \mathrm{C}$ to the water.

The equation for the fermentation of glucose to alcohol and carbon dioxide is: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ The enthalpy change for the reaction is \(-67 \mathrm{kJ} .\) Is this reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for $\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\( \)(-2755 \mathrm{kJ} / \mathrm{mol}),\( and \)\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\( calculate \)\Delta H$ for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$

Standard enthalpies of formation are relative values. What are $\Delta H_{\mathrm{f}}^{\circ}$ values relative to?

If the internal energy of a thermodynamic system is increased by $300 . \mathrm{J}\( while 75 \)\mathrm{J}$ of expansion work is done, how much heat was transferred and in which direction, to or from the system?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free