Thermochemistry

Entry Task: Block 2 Sept 27

Question:

The specific heat of copper metal is 0.385 J/g-K. How many joules of heat energy are necessary to raise the temperature of a 5.0 g block of copper

from 25.0oCto 88.5oC?

You have 5 minutes!!

Thermochemistry

Agenda:• Sign off on Ch. 5 sec 1-5 notes• HW: Calorimetric problems

Thermochemistry

I can…• Describe the relationship between energy (E),

work (w) and heat (q).• Define the state of function and how it relates to

enthalpy• Calculate the enthalpy change in a reaction• Use the calorimetry equation to solve for q.• Solve for q for solutions reactions.

Thermochemistry

Thermochemistry

Energy

• The ability to do work or transfer heat.Work: Energy used to cause an object that

has mass to move.Heat: Energy used to cause the

temperature of an object to rise.

Thermochemistry

Potential Energy

Energy an object possesses by virtue of its position or chemical composition.

Thermochemistry

Kinetic Energy

Energy an object possesses by virtue of its motion.

12

KE = mv2

Thermochemistry

Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: The calorie (cal).

1 cal = 4.184 J

1 J = 1 kg m2

s2

Thermochemistry

System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

Thermochemistry

Work

• Energy used to move an object over some distance.

• w = F d,

where w is work, F is the force, and d is the distance over which the force is exerted.

Thermochemistry

Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

Thermochemistry

First Law of Thermodynamics• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Thermochemistry

Internal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Thermochemistry

Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

Thermochemistry

Changes in Internal Energy

• If E > 0, Efinal > Einitial

Therefore, the system absorbed energy from the surroundings.

This energy change is called endothermic

Thermochemistry

Changes in Internal Energy

• If E < 0, Efinal < Einitial

Therefore, the system released energy to the surroundings.

This energy change is called exothermic.

Thermochemistry

Changes in Internal Energy

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

• That is, E = q + w.

Thermochemistry

E, q, w, and Their Signs

Thermochemistry

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

Thermochemistry

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

• When heat is released by the system to the surroundings, the process is exothermic.

Thermochemistry

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry

State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached

room temperature from either direction.

Thermochemistry

State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the

system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

Thermochemistry

State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.But q and w are different

in the two cases.

Thermochemistry

Heat (Enthalpy) change DH• Enthalpy (H) is used to quantify the heat

flow into or out of a system in a process that occurs at constant pressure.

DH = H (products) – H (reactants)

DH = heat given off or absorbed during a reaction at constant pressure

Thermochemistry

Enthalpy

Hproducts < Hreactants

DH < 0Hproducts > Hreactants

DH > 0

Thermochemistry

Endothermic and Exothermic

• A process is endothermic when H is positive.

• 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

H2O (s) H2O (l) DH = 6.01 kJ

Thermochemistry

Endothermic and Exothermic

• A process is exothermic when H is negative.

• 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ

Thermochemistry

Enthalpy of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Thermochemistry

Enthalpy of ReactionThis quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Thermochemistry

The Truth about Enthalpy1. Enthalpy is an extensive property.

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H of -890kJ

• 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l) ∆H of -1780 kJ

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

• CO2(g) + 2H2O (l) 2O2(g) + CH4(g) ∆H of +890kJ

3. H for a reaction depends on the state of the products and the state of the reactants.

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H of -890kJ

• CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆H of -802kJ

Thermochemistry

5.3 problemHydrogen peroxide can decompose to water and oxygen by the reaction

2 H2O2(l) 2 H2O(l) + O2(g) H = –196 kJ

Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure.

We have to find out how many moles of H2O2 are in 5.00g.

5.00 g 1 mole H2O2

34 g H2O2

= 0.0147 mol H2O2

0.0147 mol H2O2

2 mol H2O2

-196 kJ= -14.4 kJ

And there are 2 moles of H2O2 to release -196 kJ of energy.

Thermochemistry

5.28Consider the reaction:

CH3OH(g) CO(g) + 2H2(g) ∆H = +90.7a) Exothermic or endothermic? Its endothermic (pos ∆H)

45.0 g 1 mole CH3OH

32.04 g CH3OH

= 1.40 mol CH3OH

1.40 mol CH3OH

1 mol CH3OH

+90.7 kJ= 127 kJ

b) Calculate amount of energy when 45.0 g is transferred.

Thermochemistry

5.28Consider the reaction:

CH3OH(g) CO(g) + 2H2(g) ∆H = +90.7c) If the enthalpy is changed to 16.5 kJ, how many grams of hydrogen gas are produced?

16.5 kJ 2 mole H2

+90.7 kJ

=0.364mol H2

0.364 mol H2

1 mol H2

2 g H2 = 0.727g

Thermochemistry

5.28Consider the reaction:

CH3OH(g) CO(g) + 2H2(g) ∆H = +90.7d)How many joules of heat are released when 10.0g of CO reacts completely with hydrogen to form CH3OH at constant pressure

10.0 g CO 1 mole Co

28 g CO

=0.357mol CO

0.357 mol CO

1 mol CO

+90.7 kJ= -32.4 kJ

Thermochemistry

CalorimetrySince we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Thermochemistry

Heat Capacity and Specific HeatThe amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

Thermochemistry

Heat Capacity and Specific Heat

We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C).

Thermochemistry

Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

Specific Heat (c) =q

m T

Thermochemistry

Since we know the specific heat of most substances, we will rearrange the equation to solve for the amount of heat

(q) released or absorbed which is what we don’t know.

Thermochemistry

• q = the heat absorbed or released• c= the specific heat of water = 4.184 Joules• m= mass of the sample in grams• ΔT= the temperature difference (Tf - Ti)

q = c x m x ΔT

Thermochemistry

• q = the heat absorbed or released• c= the specific heat of ethanol= 2.44 J/g C • m= 34.4 g• ΔT= 53.8

If the temperature of 34.4g of ethanol increases from 25.0˚C to 78.8˚C, how much heat has been absorbed by

ethanol? Specific heat of ethanol is 2.44 J/g C

4515 J or 4.515 x 103 J

q=cmT

Thermochemistry

• q = 276 J• c= the specific heat of gold= 0.129 Joules• m= 4.5 g• ΔT= 25-X

A 4.50 g nugget of pure gold absorbs 276 J of heat. What was the final temperature of gold if the initial

temperature was 25.0˚C?q=cmT

276J0.129 J/(gC) x 4.5 g

= 475°C

T = q/cm =

Since the temperature started at 25.0°C, the final temperature is 500°C.

Thermochemistry

• q = 5696J• c= the specific heat = X Joules• m= 155 g• ΔT= 25-40=15

A 155 g sample of an unknown substance was heated from 25.0˚C to 40.0˚C. In the process, the substance absorbs 5696 J of energy. What is the specific heat of

the substance? What is this substance?q=cmT

5696J15 C x 155g

= 2.44 J/(gC)

c= q/Tm =

Thermochemistry

5.4 problem(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat ofthe rocks is 0.82 J/g–K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

q=cmT

a. (0.82J/g-K)(50000g)(12K) = 4.92 x 105 J

b. (0.00082 kJ/g-K)(50000g)(X) = 450 kJ

(X) = 450 kJ (41)

= 10.9 or 11.0 temp increase

Thermochemistry

WHAT ABOUT REACTIONS?

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Thermochemistry

Reactions in Solutions

Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation:

q = c m T

Thermochemistry

Reactions in SolutionsHeat produced from a reaction is

entirely absorbed by the solution and does not escape.

Exothermic rxn- heat is lost by the reaction and absorbed by the solution- temperature rises

Thermochemistry

Reactions in SolutionsHeat gained by solution, q(soln) is

therefore equal in magnitude and opposite in sign from reaction

q(soln)=

(specific heat of solution) (grams of solution)(∆T) = - q (rxn)

q(soln) = c (soln) (m soln)(∆ T) = -q(rxn)

Thermochemistry

Enthalpy of solutionEnthalpy of solution is the ∆H = qp (constant pressure)

Problem steps

1. What does solution temperature do after reaction

a. Gets warmer (solution is absorbing energy) the reaction is opposite is releasing energy ∆H is negative

b. Gets colder (solution is releasing heat) the reaction is opposite and is absorbing energy ∆H is positive.

2. Plug in your numbers qrxn=(c)(m)(∆T)

3. Change the sign to reflect concept

Thermochemistry

Enthalpy of solutionEnthalpy of solution is the ∆H = qp (constant pressure)

Problem steps

4. If it asks to calculate the enthalpy change for the reaction in kJ/mol of a substancea. Your given grams, convert to moles then divide

the ∆H by the moles.

b. Your given molarity (moles/1L = M) rearrange to get moles then divide the ∆H by the moles.

Thermochemistry

5.4 problem• When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are

mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 C to 23.11 C. The temperature increase is caused by the following reaction:

AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)

• Calculate H for this reaction in AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g C.

(0.050L)(0.100M) = 0.005 mol of AgNO3

First find the amount of energy THEN we can calculate the # of moles of a specific molarity!

q = (100g)(4.18J/g-C)(-0.81) = -338.58 J

0.005 mol of AgNO3

-338.58 J = -67716 J or -67.7 kJ

Thermochemistry

5.42 problemWhen a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.

NH4NO3(aq) NH4 +

(aq) + NO3- (aq)

• Assume that the specific heat of solution is 4.18 J/g C.

How many moles of NH4NO3 are in 4.25 g

First find the amount of energy of SOLUTION THEN we can calculate the # of moles of a specific molarity!

q = (64.25g)(4.18J/g-C)(5.1) = 1369.7 J OR 1.37 kJ

80.0 g NH4NO3

4.25 g 1 mol NH4NO3= 0.053 mol NH4NO3

Thermochemistry

5.42 problemWhen a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.

NH4NO3(aq) NH4 +

(aq) + NO3- (aq)

• Assume that the specific heat of solution is 4.18 J/g C.

So 0.053 mol NH4NO3 are in 4.25 g and 1.37 kJ of energy is released. Divide 0.053 mol into 1.37 kJ to find the number of kJ per mol.

0.053 mol of NH4NO3

1.37 kJ = 25.8 kJ or -26 kJ

Thermochemistry

5.44 problemWhen a 1.800 g sample of octane (C8H18) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/C. The temperature of the calorimeter plus contents increased from 21.36C to 28.78C What is the heat of combustion per gram of octane? Per mole of octane? 2C8H18 (g) + 17O2 (g) 16CO2

(g) + 18H2O (g)

How many moles of NH4NO3 are in 4.25 g

First find the amount of energy of SOLUTION THEN we can calculate the # of moles of a specific molarity!

q = (64.25g)(4.18J/g-C)(5.1) = 1369.7 J OR 1.37 kJ

80.0 g NH4NO3

4.25 g 1 mol NH4NO3= 0.053 mol NH4NO3

Thermochemistry

5.42 problemWhen a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.

NH4NO3(aq) NH4 +

(aq) + NO3- (aq)

• Assume that the specific heat of solution is 4.18 J/g C.

So 0.053 mol NH4NO3 are in 4.25 g and 1.37 kJ of energy is released. Divide 0.053 mol into 1.37 kJ to find the number of kJ per mol.

0.053 mol of NH4NO3

1.37 kJ = 25.8 kJ or -26 kJ

Thermochemistry

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