Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 12

When one mole of KOH is neutralized by sulfuric acid, \(q=-56 \mathrm{~kJ} .\) At \(22.8^{\circ} \mathrm{C}, 25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is neutralized by \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) \(\mathrm{KOH}\) in a coffee-cup calorimeter. What is the final temperature of the solution? (Use the assumptions in Question 11.)

Short Answer

Expert verified
Answer: The final temperature of the solution is 27.27°C.

Step by step solution

01

Determine the moles of limiting reactant

To find out which reactant is limiting, examine the moles of each reactant: Moles of H₂SO₄ = Volume × Concentration = (25 mL) × (0.5 mol/L) = 0.0125 mol Moles of KOH = Volume × Concentration = (50 mL) × (0.5 mol/L) = 0.025 mol Since KOH reacts with H₂SO₄ in a 2:1 ratio, the amount of KOH required for 0.0125 mol of H₂SO₄ is 0.025 mol which is exactly the available amount of KOH. Therefore, KOH is the limiting reactant.
02

Calculate the heat change during the reaction

The heat change (q) per mole of KOH during the neutralization is given as -56 kJ. Since all 0.025 mol of KOH will be reacting, the total heat change produced by the reaction is: Total heat change, q = moles × heat change per mole = 0.025 mol × (-56 kJ/mol) = -1.4 kJ
03

Calculate the mass of the solution

The total volume of the solution is given by adding the volume of H₂SO₄ and KOH: Total volume = 25 mL + 50 mL = 75 mL Assuming 1 g/mL as the density of the solution (which is very close to the density of water), we can find the mass of the solution: Mass = Volume × Density = 75 mL × 1 g/mL = 75 g
04

Calculate temperature change

We are given that the calorimeter is well-insulated, so we can assume that all the heat released by the reaction is absorbed by the solution. To calculate the temperature change of the solution, we can use the formula: ΔT = q / (mass × C), where ΔT is the temperature change, q is the heat change, mass is the mass of the solution, and C is the specific heat of water (4.18 J/gC). Since the heat change is negative (exothermic reaction), the temperature will increase. Convert q to J by multiplying with 1000: -q = 1.4 kJ × 1000 J/kJ = 1400 J ΔT = 1400 J / (75 g × 4.18 J/gC) = 4.47 °C
05

Calculate final temperature

The initial temperature of the solution is given as 22.8°C. The final temperature of the solution can be calculated by adding the temperature change to the initial temperature: Final temperature = Initial temperature + ΔT = 22.8°C + 4.47°C = 27.27°C Therefore, the final temperature of the solution is 27.27°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.

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

A sample of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), is contaminated by sodium chloride. When the contaminated sample is burned in a bomb calorimeter, sodium chloride does not burn. What is the percentage of sucrose in the sample if a temperature increase of \(1.67^{\circ} \mathrm{C}\) is observed when \(3.000 \mathrm{~g}\) of the sample is burned in the calorimeter? Sucrose gives off \(5.64 \times 10^{3} \mathrm{~kJ} / \mathrm{mol}\) when burned. The heat capacity of the calorimeter is \(22.51 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\).

Calculate (a) \(g\) when a system does \(54 \mathrm{~J}\) of work and its energy decreases by \(72 \mathrm{~J}\). (b) \(\Delta E\) for a gas that releases \(38 \mathrm{~J}\) of heat and has \(102 \mathrm{~J}\) of work done on it.

On a hot day, you take a six-pack of soda on a picnic, cooling it with ice. Each empty (aluminum) can weighs \(12.5 \mathrm{~g}\) and contains \(12.0\) oz of soda. The specific heat of aluminum is \(0.902 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\); take that of soda to be \(4.10 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) (a) How much heat must be absorbed from the six-pack to lower the temperature from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\) ? (b) How much ice must be melted to absorb this amount of heat? \(\left(\Delta H_{\mathrm{fus}}\right.\) of ise is given in Table 8.2.)

Copper is used in building the integrated circuits, chips, and printed circuit boards for computers. When \(228 \mathrm{~J}\) of heat are absorbed by \(125 \mathrm{~g}\) of copper at \(22.38^{\circ} \mathrm{C}\), the temperature increases to \(27.12^{\circ} \mathrm{C}\). What is the specific heat of copper?

Titanium is a metal used in jet engines. Its specific heat is \(0.523 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If \(5.88 \mathrm{~g}\) of titanium absorbs \(4.78 \mathrm{~J}\), what is the change in temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Nuclear 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