Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 60

Under constant-volume conditions, the heat of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) ) is 26.38 \(\mathrm{kJ} / \mathrm{g} .\) A 2.760 -g sample of \right. benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.60 to \(29.93^{\circ} \mathrm{C}\) (a) What is the total heat capacity of the calorimeter? \(\mathrm{b}\) ) \(\mathrm{A} 1.440\) -g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from 22.14 to \(27.09^{\circ} \mathrm{C} .\) What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

Short Answer

Expert verified
The total heat capacity of the calorimeter is 8.74 kJ/°C. The heat of combustion per gram of the new substance is 30.04 kJ/g. If a portion of water in the calorimeter is lost, the heat capacity of the calorimeter would decrease, possibly affecting the measured change in temperature and the accuracy of calculations.

Step by step solution

01

Calculate the heat of combustion for the benzoic acid sample

The heat of combustion of benzoic acid is given as 26.38 kJ/g. To find the heat of combustion for the 2.760 g sample, we can use the following formula: Heat = mass × heat of combustion per gram \(q_{benzoic\_acid} = m \times H\) where q is the heat, m is the mass, and H is the heat of combustion per gram. \(q_{benzoic\_acid} = 2.760 \times 26.38 = 72.8072 kJ\)
02

Calculate the total heat capacity of the calorimeter

We can use the change in temperature for the calorimeter to calculate its total heat capacity. The change in temperature is the final temperature minus the initial temperature: ΔT = T_final - T_initial ΔT = \(29.93^{\circ} \mathrm{C} - 21.60^{\circ} \mathrm{C} = 8.33^{\circ} \mathrm{C}\) Using the heat of the benzoic acid sample and the change in temperature, we can find the total heat capacity C of the calorimeter: C = q/ΔT C = 72.8072 kJ / 8.33 °C C = 8.74 kJ/°C
03

Calculate the heat transfer of the new substance

Next, we can find the heat transfer for the new substance using the change in temperature of the calorimeter: ΔT = T_final - T_initial ΔT = \(27.09^{\circ} \mathrm{C} - 22.14^{\circ} \mathrm{C} = 4.95^{\circ} \mathrm{C}\) Now, using the total heat capacity of the calorimeter found in step 2 and the change in temperature, we can find the heat transfer of the new substance: q = C × ΔT q = 8.74 kJ/°C × 4.95 °C q = 43.261 kJ
04

Calculate the heat of combustion per gram of the new substance

Now that we have the heat transfer of the new substance, we can find the heat of combustion per gram. We are given that the mass of the new substance is 1.440 g: Heat of combustion per gram (H) = q / m H = 43.261 kJ / 1.440 g H = 30.04 kJ/g
05

Discuss the effect of water loss on the calorimeter's heat capacity

In case a portion of water in the calorimeter is lost, the heat capacity of the calorimeter would decrease. This is because water has a specific heat capacity and, as such, contributes to the overall heat capacity of the calorimeter. If less water is present, the calorimeter will absorb less heat, resulting in a lower heat capacity. However, this would not directly affect the heat of combustion of either substance, though it may affect the measured change in temperature and result in less accurate calculations.

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

The Sun supplies about 1.0 kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{kW} / \mathrm{m}^{2},\) where a watt \(=1 \mathrm{J} / \mathrm{s}\right)\) Plants produce the equivalent of about 0.20 \(\mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$\begin{array}{c}{12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\Delta H=5645 \mathrm{kJ}}\end{array}$$

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l) :\) $$\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H &=-726.5 \mathrm{kJ} \end{aligned}$$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: $$\begin{array}{l}{\text { (a) } \mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(\mathrm{g})} \\ {\text { (b) } 2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g)} \\ {\text { (c) } \mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}} \\ {\text { (d) } \mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)}\end{array}$$

At the end of \(2012,\) global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H^{\circ}=&-2803 \mathrm{kJ} \end{aligned}$$

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

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