Energy Capacity to do work or to produce heat.

Energy Capacity to do work or to produce heat.
Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. The total energy content of the universe is constant. Copyright © Cengage Learning. All rights reserved

Energy Potential energy – energy due to position or composition.
Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Copyright © Cengage Learning. All rights reserved

Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Copyright © Cengage Learning. All rights reserved

Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work – force acting over a distance. Energy is a state function; work and heat are not State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Copyright © Cengage Learning. All rights reserved

Chemical Energy Endothermic Reaction: Heat flow is into a system.
Absorb energy from the surroundings. Exothermic Reaction: Energy flows out of the system. Energy gained by the surroundings must be equal to the energy lost by the system. Copyright © Cengage Learning. All rights reserved

Endothermic and Exothermic Systems and Energy

CONCEPT CHECK! Is the freezing of water an endothermic or exothermic process? Explain. Exothermic process because you must remove energy in order to slow the molecules down to form a solid. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Exo Endo Exothermic (heat energy leaves your hand and moves to the ice) Endothermic (heat energy flows into the ice) Endothermic (heat energy flows into the water to boil it) Exothermic (heat energy leaves to condense the water from a gas to a liquid) Endothermic (heat energy flows into the ice cream to melt it) Copyright © Cengage Learning. All rights reserved

Methane is burning in a Bunsen burner in a laboratory.
CONCEPT CHECK! For each of the following, define a system and its surroundings and give the direction of energy transfer. Methane is burning in a Bunsen burner in a laboratory. Water drops, sitting on your skin after swimming, evaporate. System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic) System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic) Copyright © Cengage Learning. All rights reserved

Hydrogen gas and oxygen gas react violently to form water. Explain.
CONCEPT CHECK! Hydrogen gas and oxygen gas react violently to form water. Explain. Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted. Copyright © Cengage Learning. All rights reserved

Thermodynamics The study of energy and its interconversions is called thermodynamics. Law of conservation of energy is often called the first law of thermodynamics. Copyright © Cengage Learning. All rights reserved

Internal Energy Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system. To change the internal energy of a system: ΔE = q + w q represents heat w represents work Copyright © Cengage Learning. All rights reserved

Internal Energy Thermodynamic quantities consist of two parts:
Number gives the magnitude of the change. Sign indicates the direction of the flow. Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view.
Endothermic Process: q is positive Exothermic Process: q is negative Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view.
System does work on surroundings: w is negative Surroundings do work on the system: w is positive Copyright © Cengage Learning. All rights reserved

Work Work = P × A × Δh = PΔV P is pressure. A is area.
Δh is the piston moving a distance. ΔV is the change in volume.

They perform the same amount of work.
EXERCISE! Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. They both perform the same amount of work. w = -PΔV a) w = -(2 atm)( ) = -6 L·atm b) w = -(3 atm)( ) = -6 L·atm Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Determine the sign of ΔE for each of the following with the listed conditions: a) An endothermic process that performs work. |work| > |heat| |work| < |heat| b) Work is done on a gas and the process is exothermic. Δ E = negative Δ E = positive a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative Copyright © Cengage Learning. All rights reserved

Change in Enthalpy State function ΔH = q at constant pressure
ΔH = Hproducts – Hreactants Copyright © Cengage Learning. All rights reserved

–252 kJ Consider the combustion of propane: EXERCISE! ΔH = –2221 kJ
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. –252 kJ (5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8) ΔH = -252 kJ Copyright © Cengage Learning. All rights reserved

Calorimetry Science of measuring heat Specific heat capacity:
The energy required to raise the temperature of one gram of a substance by one degree Celsius. Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved

Calorimetry If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution. Copyright © Cengage Learning. All rights reserved

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups
Copyright © Cengage Learning. All rights reserved

Calorimetry Energy released (heat) = s × m × ΔT
s = specific heat capacity (J/°C·g) m = mass of solution (g) ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A g sample of water at 90°C is added to a g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C The correct answer is b). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A g sample of water at 90.°C is added to a g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C The correct answer is c). The final temperature of the water is 23°C. - (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 23°C Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C The correct answer is c). The final temperature of the water is 18°C. - (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 18°C Copyright © Cengage Learning. All rights reserved

Interactive Example 6.5 - Constant-Pressure Calorimetry
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a calorimeter, the white solid BaSO4 forms, and the temperature of the mixture increases to 28.1°C

Interactive Example 6.5 - Constant-Pressure Calorimetry (Continued)
Assume that: The calorimeter absorbs only a negligible quantity of heat The specific heat capacity of the solution is 4.18 J/ °C · g The density of the final solution is 1.0 g/mL Calculate the enthalpy change per mole of BaSO4 formed

Interactive Example 6.5 - Solution
Where are we going? To calculate ΔH per mole of BaSO4 formed What do we know? 1.00 L of 1.00 M Ba(NO3)2 1.00 L of 1.00 M Na2SO4 Tinitial = 25.0°C and Tfinal = 28.1°C Heat capacity of solution = 4.18 J/ °C · g Density of final solution = 1.0 g/mL

Interactive Example 6.5 - Solution (Continued 1)
What do we need? Net ionic equation for the reaction The ions present before any reaction occurs are Ba2+, NO3–, Na+, and SO42– The Na+ and NO3– ions are spectator ions, since NaNO3 is very soluble in water and will not precipitate under these conditions The net ionic equation for the reaction is:

How do we get there? What is ΔH? Since the temperature increases, formation of solid BaSO4 must be exothermic ΔH is negative Heat evolved by the reaction = heat absorbed by the solution = specific heat capacity×mass of solution×increase in temperature

What is the mass of the final solution? What is the temperature increase? How much heat is evolved by the reaction?

Thus, What is ΔH per mole of BaSO4 formed? Since 1.0 L of 1.0 M Ba(NO3)2 contains 1 mole of Ba2+ ions and 1.0 L of 1.0 M Na2SO4 contains 1.0 mole of SO42– ions, 1.0 mole of solid BaSO4 is formed in this experiment Thus the enthalpy change per mole of BaSO4 formed is

Constant-Volume Calorimetry
Used in conditions when experiments are to be performed under constant volume No work is done since V must change for PV work to be performed Bomb calorimeter Weighed reactants are placed within a rigid steel container and ignited Change in energy is determined by the increase in temperature of the water and other parts

Figure 6.6 - A Bomb Calorimeter

Constant-Volume Calorimetry (Continued)
For a constant-volume process, ΔV = 0 Therefore, w = –PΔV = 0 Energy released by the reaction = temperature increase ×energy required to change the temperature by 1°C = ΔT×heat capacity of the calorimeter

Hess’s Law In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps Copyright © Cengage Learning. All rights reserved 41 41

Characteristics of Enthalpy Changes
If a reaction is reversed, the sign of ΔH is also reversed Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer Copyright © Cengage Learning. All rights reserved 42 42

Problem-Solving Strategy - Hess’s Law
Work backward from the required reaction Use the reactants and products to decide how to manipulate the other given reactions at your disposal Reverse any reactions as needed to give the required reactants and products Multiply reactions to give the correct numbers of reactants and products Copyright © Cengage Learning. All rights reserved 43 43

Interactive Example 6.8 - Hess’s Law II
Diborane (B2H6) is a highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program Calculate ΔH for the synthesis of diborane from its elements, according to the following equation:

Interactive Example 6.8 - Hess’s Law II (Continued)
Use the following data: Reaction ΔH –1273 kJ –2035 kJ –286 kJ 44 kJ

To obtain ΔH for the required reaction, we must somehow combine equations (a), (b), (c), and (d) to produce that reaction and add the corresponding ΔH values This can best be done by focusing on the reactants and products of the required reaction The reactants are B(s) and H2(g), and the product is B2H6(g)

How can we obtain the correct equation? Reaction (a) has B(s) as a reactant, as needed in the required equation Reaction (a) will be used as it is Reaction (b) has B2H6(g) as a reactant, but this substance is needed as a product Reaction (b) must be reversed, and the sign of ΔH must be changed accordingly

Up to this point we have: Deleting the species that occur on both sides gives: ΔH = 762 kJ

We are closer to the required reaction, but we still need to remove H2O(g) and O2(g) and introduce H2(g) as a reactant We can do this using reactions (c) and (d) Multiply reaction (c) and its ΔH value by 3 and add the result to the preceding equation

ΔH = –96 kJ

We can cancel the 3/2 O2(g) on both sides, but we cannot cancel the H2O because it is gaseous on one side and liquid on the other This can be solved by adding reaction (d), multiplied by 3: ΔH = +36 kJ

This gives the reaction required by the problem Conclusion ΔH for the synthesis of 1 mole of diborane from the elements is +36 kJ

Standard Enthalpy of Formation (ΔHf°)
Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Copyright © Cengage Learning. All rights reserved

Conventional Definitions of Standard States
For a Compound For a gas, pressure is exactly 1 atm. For a solution, concentration is exactly 1 M. Pure substance (liquid or solid) For an Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C. Copyright © Cengage Learning. All rights reserved

ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ
A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH°reaction = –(–75 kJ) (–394 kJ) + (–572 kJ) = –891 kJ Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy: Enthalpy Calculations
When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer. Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy: Enthalpy Calculations
The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:  H°rxn =  npHf°(products) -  nr  Hf°(reactants) 4. Elements in their standard states are not included in the ΔHreaction calculations because ΔHf° for an element in its standard state is zero. Copyright © Cengage Learning. All rights reserved

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:
EXERCISE! Calculate  H° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:  Hf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 ΔH° = –368 kJ [2(–470) + 0] – [0 + 2(–286)] = –368 kJ ΔH = –368 kJ Copyright © Cengage Learning. All rights reserved

Interactive Example 6.10 - Enthalpies from Standard Enthalpies of Formation II
Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction: This reaction occurs when a mixture of powdered aluminum and iron(III) oxide is ignited with a magnesium fuse

Where are we going? To calculate ΔH for the reaction What do we know? ΔHf°for Fe2O3(s) = – 826 kJ/mol ΔHf°for Al2O3(s) = – 1676 kJ/mol ΔHf°for Al(s) = 0 for Fe(s) = 0

What do we need? We use the following equation: How do we get there?

This reaction is so highly exothermic that the iron produced is initially molten Used as a lecture demonstration Used in welding massive steel objects such as ships’ propellers

Interactive Example 6.13 - Comparing Enthalpies of Combustion
Assuming that the combustion of hydrogen gas provides three times as much energy per gram as gasoline, calculate the volume of liquid H2 (density = g/mL) required to furnish the energy contained in 80.0 L (about 20 gal) of gasoline (density = g/mL) Calculate also the volume that this hydrogen would occupy as a gas at 1.00 atm and 25°C

Where are we going? To calculate the volume of H2(l) required and its volume as a gas at the given conditions What do we know? Density for H2(l) = g/mL 80.0 L gasoline Density for gasoline = g/mL H2(g) ⇒ P = 1.00 atm, T = 25°C = 298 K

How do we get there? What is the mass of gasoline? How much H2(l) is needed? Since H2 furnishes three times as much energy per gram as gasoline, only a third as much liquid hydrogen is needed to furnish the same energy

Since density = mass/volume, then volume = mass/density, and the volume of H2(l) needed is Thus, 277 L of liquid H2 is needed to furnish the same energy of combustion as 80.0 L of gasoline

What is the volume of the H2(g)? To calculate the volume that this hydrogen would occupy as a gas at 1.00 atm and 25°C, we use the ideal gas law: In this case: P = 1.00 atm T = °C = 298 K R = L · atm/K · mol

What are the moles of H2(g)? Thus, At 1 atm and 25°C, the hydrogen gas needed to replace 20 gal of gasoline occupies a volume of 239,000 L

Energy Capacity to do work or to produce heat.

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