/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Consider the reaction $$ 2 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 70

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

Short Answer

Expert verified
The final temperature of the mixture when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.100 M Ba(OH)鈧� is approximately 27.12掳C.

Step by step solution

01

Calculate the limiting reagent

Determine the limiting reagent by calculating the moles of HCl and Ba(OH)鈧� in the given volumes and concentrations of each solution. Moles of HCl: moles = Molarity 脳 Volume moles = 0.500 M 脳 0.100 L = 0.050 moles Moles of Ba(OH)鈧�: moles = Molarity 脳 Volume moles = 0.100 M 脳 0.300 L = 0.030 moles Based on the stoichiometry of the reaction (2 moles of HCl react with 1 mole of Ba(OH)鈧�), twice the moles of HCl are needed in comparison to the moles of Ba(OH)鈧�. But in this case, there are more moles of HCl present than required, making Ba(OH)鈧� the limiting reagent.
02

Calculate the heat evolved in the reaction

Using the moles of the limiting reactant (Ba(OH)鈧�), the enthalpy of the reaction (螖H), and the stoichiometry of the reaction, calculate the heat evolved (q) during the reaction. q (reaction) = Moles of limiting reactant 脳 螖H q (reaction) = 0.030 moles Ba(OH)鈧� 脳 (-118000 J/mol) q (reaction) = -3540 J Notice that the result is negative, which indicates that the system has released heat (exothermic reaction).
03

Calculate the heat absorbed by the solution

Now, determine the heat absorbed by the solution (q(solution)), which is equal in magnitude but opposite in sign to the heat evolved in the reaction (q(reaction)). Calculate the temperature change in the solution using this heat absorbed and the specific heat capacity provided. q (solution) = -q (reaction) q (solution) = 3540 J 螖T (temperature change) = q(solution) / (mass 脳 specific heat capacity) 螖T = 3540 J / (400.0 g 脳 4.18 J/g路掳C) 螖T = 2.12掳C
04

Calculate the final temperature of the mixture

Using the initial temperature (25.0掳C) and the temperature change (螖T) calculated in the previous step, determine the final temperature of the mixture. Final temperature = Initial temperature + 螖T Final temperature = 25.0掳C + 2.12掳C Final temperature = 27.12掳C Now we have determined that when mixing 100 mL of 0.500 M HCl with 300 mL of 0.100 M Ba(OH)鈧�, the final temperature of the resulting solution is approximately 27.12掳C.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reagent
In any chemical reaction, the limiting reagent is the reactant that is completely consumed first. This reactant sets a limit on the amount of products that can be formed. To find the limiting reagent, first calculate the number of moles for each reactant based on the given concentration and volume.
For the reaction between HCl and Ba(OH)鈧�, we calculate:
  • Moles of HCl: 0.500 M 脳 0.100 L = 0.050 moles
  • Moles of Ba(OH)鈧�: 0.100 M 脳 0.300 L = 0.030 moles
The balanced equation shows that 2 moles of HCl react with 1 mole of Ba(OH)鈧�. Therefore, for 0.030 moles of Ba(OH)鈧�, we need 0.060 moles of HCl, but only 0.050 moles are available. Hence, Ba(OH)鈧� is the limiting reagent.
Understanding which reactant is the limiting reagent helps us predict how much product can be formed and how much heat will be released in an exothermic reaction.
Enthalpy Change
Enthalpy change (\( \Delta H \)) indicates whether a reaction releases or absorbs energy when products are formed from reactants. It forms a crucial part of understanding the heat involved in chemical reactions. For our reaction:
  • The reaction has an enthalpy change of \( \Delta H = -118 \ \text{kJ/mol} \)
The negative sign shows that the reaction is exothermic, releasing heat to the surroundings.
Calculating the heat released involves using the limiting reagent to determine the amount of energy change:
  • Heat released (\( q \)) = Moles of limiting reagent 脳 \( \Delta H \)
  • For Ba(OH)鈧�: \( q \) = 0.030 moles 脳 (-118,000 J/mol) = -3540 J
This calculation reveals the amount of heat released due to the reaction, and since the result is negative, it confirms that the reaction is exothermic. This released heat can also result in a temperature change of the reacting system.
Exothermic Reaction
An exothermic reaction is one that releases heat, resulting in an increase in the temperature of the surroundings. In such reactions, the energy required to break the bonds in reactants is less than energy released when new bonds form in the products. In this particular case, the reaction between HCl and Ba(OH)鈧� is exothermic, evident from the negative enthalpy change ( \( \Delta H = -118 \ \text{kJ/mol} \)).
As per the step-by-step method, when \(0.030\) moles of Ba(OH)鈧� react, \(3540\) J of heat is released.
The temperature of the solution increases as a result of this heat release.
  • Initial temperature: \(25.0^{\circ} \text{C}\)
  • Temperature change: \(2.12^{\circ} \text{C}\)
  • Final temperature: \(27.12^{\circ} \text{C}\)
The increase in temperature is a direct consequence of the exothermic nature of the reaction.

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

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

A balloon filled with 39.1 moles of helium has a volume of 876 \(\mathrm{L}\) at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of 998 \(\mathrm{L}\) , the pressure remaining constant. Calculate \(q, w,\) and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{mol.} )\)

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at 298 \(\mathrm{K}\) and \(1.0 \mathrm{atm},\) \(\Delta H\) is more positive than \(\Delta E\) by 2.5 \(\mathrm{kJ} / \mathrm{mol}\) . What does the 2.5 \(\mathrm{kJ} / \mathrm{mol}\) quantity represent?

A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and 1.00 atm pressure. \(\Delta H_{\text { vap }}\) for the liquid is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C} .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

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.