Download Energy and Chemical Reactions - Thermochemistry

Energy and Chemical Reactions - Thermochemistry
Alkali Metals in Water: http://www.youtube.com/watch?v=uixxJtJPVXk&feature=related
Key Concepts-Our Goals are to:
1.
develop an understanding of Potential and Kinetic Energy and the ability to describe the relationship
between the two in Thermochemistry.
2.
be able to define the terms System and Surroundings and to identify a System and its Surroundings
for physical and chemical changes.
3.
develop an understanding of the concept of Heat Transfer and the ability to describe the process.
4.
develop an understanding of the concept of Pressure-Volume Work and the ability to recognize when
a system can do work or when a system has work done on it.
5.
develop the ability to apply and use the First Law of Thermodynamics.
6.
develop an understanding of internal energy change for a system, ∆Esys; be able to relate this to
heat flow and work (Learn the sign conventions!).
7.
develop an understanding of the concept of a State Function; know what variables in chemistry are
State Functions.
8.
develop an understanding of ENTHALPY Change, ∆H; know that this is heat transfer at constant
pressure, a State Function.
9.
develop an understanding of Enthalpies of Reaction, ∆Hrxn; be able to apply Hess’s Law to determine
∆Hrxn and to calculate ∆Hrxn from measured heat flow.
10. develop an understanding of Enthalpies of Formation, ∆H°f, and the ability to write reactions
corresponding to standard enthalpies of formation.
11. develop the ability to do stoichiometric calculations that include heats of reaction.
12. learn the basics of Calorimetry; be able to use calorimetry data to calculate ∆Hrxn, specific heat
capacity, temperature change, amount of heat flow, etc..(Covered in LAB only.)
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Energy-Kinetic vs. Potential Energy
Energy - The capacity to do work. Common units are Joules and Calories
Joule (J): SI unit of energy (A very small amount of energy so we commonly use the kilojoule, kJ.)
1 J = 1 Newton•m = 1 kg•m2/s2 calorie (cal): 1 cal = 4.184 J (The heat required to raise the temperature of 1 g of water from 14.5°C to 1 5.5°C.)
Nutritional calorie (Cal) = 1 kcal = 1000 cal
Kinetic Energy: Energy due to motion of matter.
In chemistry we are often concerned with the thermal (heat) energy of atoms/molecules/ions. Heat is a
manifestation of random kinetic energy. Heat is directly proportional to temperature.
Heat and Temperature are not the same!
Heat is a measure of energy content.
Heat is an extensive property.
Temperature is a measure of AVERAGE kinetic energy.
Temperature is an intensive property.
Heat is always transferred from an object with a higher temperature
(higher average kinetic energy of the particles) to one of lower
temperature (lower average kinetic energy of the particles) until
THERMAL EQUILIBRIUM is reached.
Amount of Heat Lost = Amount of Heat Gained
Potential Energy: Stored energy in a system. Weakly temperature dependent.
Chemical Potential Energy-The kind of Potential Energy of interest in Thermochemistry.
Chemical potential energy results from attractive and repulsive forces between nuclei and electrons in a system.
Includes the bond energies between atoms bonded together within a molecule (intramolecular), forces between
ions, and forces between different molecules (intermolecular) in liquids and solids.
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Kinetic Energy
Kinetic Energy (KE)
To calculate the kinetic energy of a moving object in joules, what units must you use for the
object’s mass, m, and velocity, u?
KE =
1 2
mu
2
Text Question 5.16:
a) A baseball weighs 5.13 oz. What us the kinetic energy in joules of this baseball when it
is thrown by a major-league pitcher at 95.0 mph?
b)
By what factor will the kinetic energy change if the speed of the baseball is decreased to
55.0 mph?
c)
What happens to the kinetic energy when the baseball is caught by the catcher?
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Thermochemistry Basic Terms
When we study chemical systems we actually study ENERGY TRANSFER NOT ENERGY CONTENT.
Energy is TRANSFERRED between a SYSTEM and SURROUNDINGS as heat (q) and/or work (w).
System: Process under study (chemical reaction or physical change).
Surroundings: Everything else. May be isolated as in a calorimetry experiment.
Closed System: A system that can exchange energy, but not matter with its surroundings.
Universe = System + Surroundings
System
Universe
heat
q
work
w
Surroundings
The Internal Energy (E) of the SYSTEM: The sum of all the kinetic and potential energies
of the matter in the system. CANNOT BE absolutely determined.
However, we can easily measure changes in internal energy (∆Esys) for a system:
∆Esys = Efinal - Einitial = q + w
∆ Esys (+) SYSTEM internal energy INCREASES.
∆ Esys (−) SYSTEM internal energy DECREASES.
Sign conventions:
1. q (+) HEAT flows into the system from the surroundings. ENDOTHERMIC PROCESS
2. q (-) HEAT flows out of the system to the surroundings. EXOTHERMIC PROCESS
3. w(+) WORK is done ON the system by the surroundings.
4. w(-) WORK is done BY the system on the surroundings.
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First Law of
Thermodynamics:
Energy can be transferred
from one form to another,
but cannot be created or
destroyed in a physical or
chemical change.
∆ Euniv = ∆ Esys + ∆ Esurr = 0
∆ Esys = −∆ Esurr
Definitions and Sign Conventions for Work (w)
Work (w): Energy that moves an object against a force is called work. Work is quantified as
force x distance:
w = F •d
Only pressure-volume work will be considered here. For a gaseous system expanding or
contracting in volume (distance) of ∆V = Vfinal – Vinital, against a constant external pressure
(force), P, the work is defined as:
w = F • d = −P(∆ V )
SI units are pascals for pressure and m3 for volume. Chemists like to use atmospheres for
pressure and liters for volume. To convert to Joules we use 1 L•atm = 101.325 J.
wsys > 0, ∆Esys increases
Work is done on the system. This is the case when the volume of a gas system decreases,
∆V is negative, so w = –P∆V is positive.
wsys < 0, ∆Esys decreases
Work is done by the system. This is the case when the volume of a gas system increases, ∆V
is positive, so w = –P∆V is negative.
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Example Work for a Physical Change: Solid —> Gas
Reaction systems that result in a change in moles of gas experience a volume change. An example
where work is done by the system is sublimation of CO2: CO2(s) —> CO2(g)
Another example is the denotation of TNT.
Here 15 moles of gas are created for every
2 moles (450 g) of TNT. Assuming a
temperature of 525°C, the hot expanding
gasses can do ≈100 kJ of work.
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Can you think of a simple example where wsys > 0?
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State Functions in Chemistry
State functions are functions whose CHANGE in value depends only
upon the the initial and final states of the system, not on the pathway
from initial to final.
∆(State function) = Final State - Initial State
By convention, all state functions are represented by UPPERCASE letters
in chemistry.
∆T (temperature), ∆P (pressure), ∆V (volume), ∆E (internal energy) are
all state functions.
Variables that are not state functions: w (work), d (distance), t (time),
q (heat), etc.
Note: the sum q + w = ∆E is a state function.
A Special State Function: ENTHALPY, H
At constant external pressure, ∆P = 0, the heat transferred
between the system and surroundings is called enthalpy (H). The
change in enthalpy is defined as:
∆ H sys = ∆ E p + P ∆V = ?
∆Hsys is the Enthalpy of Reaction
(Heat of reaction at constant P, a STATE FUNCTION.)
The relative amounts of heat and work for a
given ∆E depends upon the pathway over
which the energy change occurs. Consider the
discharge of the potential energy stored in a
battery. We can vary the proportions of heat to
work by changing the circuitry attached to the
battery. The same amount of energy is used in
both cases, but the pathway of energy use is
different.
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Heat Transfer at Constant Pressure, ∆H
Note how changes in internal energy and changes in enthalpy (heat content) at constant pressure
are related, differing only in the quantity of energy transferred to or from the system as work:
∆E = ∆H – P∆V = ∆H + w
(Remember w = – P∆V)
For many processes ∆V is very small, or negligible, making ∆E ≈ ∆H.
What are some examples of processes with negligible ∆V?
(Hint: What processes DO NOT involve a change in the number of moles of gas, thus ∆V ≈ 0?)
In processes where ∆V is large, ∆E and ∆H will be different.
What are some examples of processes with significant ∆V?
(Hint: What processes DO involve a change in the number of moles of gas, thus ∆V ≠ 0?)
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Example Chemical System
Consider the case where this reaction is done at a
constant external pressure of 1.00 atm and T = 25°C.
For the complete reaction of 1 mole of zinc:
1.
The amount of heat absorbed by the surroundings
is 154 kJ.
2.
Zn(s) + 2 HCl(aq)
SYSTEM
Enthalpy, ∆Hsys
Consider the redox reaction of zinc metal
with hydrochloric acid:
Zn(s) + 2 HCl(aq) –> ZnCl2(aq) + H2(g)
Initial State of
SYSTEM
Enthalpy Diagram
–154 kJ
ZnCl2(aq) + H2(g)
Final State of
SYSTEM
The volume of H2 gas generated is 24.5 liters.
We symbolize and quantify this as follows:
1. qp = ∆Hsys = –154 kJ (Exothermic)
Work
2. ∆Hsurr = – ∆Hsys = +154 kJ
3. w = -P(∆V) = (-1.00 atm)(+24.5 L)
= -24.5 L*atm = -2.48 kJ
4. ∆Esys = ∆Hsys + w = (–154 + -2.48) kJ = -156 kJ
5. ∆Esurr = – ∆Esys = +156 kJ
Observations:
The chemical reaction is EXOTHERMIC. The surroundings will gain heat and should show an increase in T.
The SYSTEM loses 154 kJ of POTENTIAL energy (not thermal energy) to the surroundings.
The SYSTEM does 2.48 kJ of work on the surroundings.
In this chemical system, |∆Hsys| >> |w| making ∆Esys ≈ ∆Hsys.
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∆H of Reactions, ∆Hsys
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(l) ∆Hrxn = –890 kJ
For the above reaction the value of ∆Hsys represents the
difference in chemical potential energy between the
products (final state) and the reactants (initial state)
∆Hsys = H(products) - H(reactants)
So if ∆Hsys is a difference in potential energy then why do we
refer to it as a heat? Because we measure the system’s
difference in potential energy (∆Hsys) by measuring the HEAT
change of the surroundings (qsurr) when the chemical
reaction takes place.
∆(potential energysys) = ∆(heatsurr)
Thus, ∆Hsys = -qsurr. In the surroundings we measure a heat
change by experimentally observing a change in temperature.
This is the basis for calorimetry (covered in lab).
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Review of Energy Changes
Universe
Potential Energy
Changes
Kinetic Energy
Changes
(Temperature constant)
heat
work
System
(Temperature changes)
q
w
Surroundings
Make sure that you understand the changes in the system versus the surroundings when we study chemical
reactions and physical state changes. The “system” undergoes potential energy changes when the reactants
are transformed into products or when a physical state change occurs. The temperature of the system is taken
as the same before and after reaction, no thermal energy change in the system. Heat flow to or from the
surroundings maintains the system’s temperature.
The surroundings are a reservoir of thermal energy where heat is absorbed or released. Any potential energy
change in the system is reflected by a corresponding thermal energy change in the surroundings. This is what
we can “see” (measure) in the laboratory. How do we know when a potential energy change takes place in a
system? We observe the effects of the system’s potential energy change on the surroundings because the
surroundings experience a change in temperature. This is the basis for calorimetry.
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Energy of Reactions
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Heat of Reactions, ∆Hrxn
For chemical reactions we measure or calculate ∆Hsys for a balanced chemical reaction.
CH4(g) + 2O2(g) —> CO2(g) + 2 H2O(l) ∆Hrxn = –890 kJ = ∆Hsys
1.
∆Hrxn is specific to the reaction given with the coefficients given.
2.
The physical states (s, l, g, aq) are required and important!
3.
The coefficients in the balanced equation determine the extent of ∆Hrxn.
∆Hrxn therefore depends on the amount of reaction, and is an EXTENSIVE property.
If you multiply the coefficients of a balanced equation by a constant, ∆Hrxn is
multiplied by the same constant.
4.
If a chemical reaction is reversed, then the sign of ∆Hrxn changes but the magnitude
remains the same.
CO2(g) + 2H2O(l) —> CH4(g) + 2 O2(g) ∆Hrxn = +890 kJ
5.
For an exothermic system we think of heat as a product of the reaction:
reactants —> products + heat
(Thermodynamically favorable, but can still be non-spontaneous for certain systems under certain conditions.
This will be explored in more detail in chemistry 1B.)
6.
For an endothermic system we think of heat as a reactant of the reaction:
heat + reactants —> products
(Thermodynamically unfavorable, but can still occur spontaneously for certain systems under the right
conditions. This will be explored in more detail in chemistry 1B.)
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Stoichiometry and ∆Hrxn
For the reaction:
N2(g) + 3 H2(g) —> 2 NH3(g)
∆Hrxn = –91.80 kJ (so heat is a product, exothermic)
We can interpret the magnitude of ∆Hrxn in many ways:
1. 91.80 kJ of heat is produced when 1 mol of N2 reacts or
2. 91.80 kJ of heat is produced when 3 mol of H2 reacts or
3. 91.80 kJ of heat is produced when 2 mol of NH3 is formed
1.
What is ∆H/mol N2(g), ∆H/mol H2(g), ∆H/mol NH3(g)?
2.
How much heat is produced if 23.6 g of hydrogen reacts?
3.
How many grams of ammonia are produced when 750 kJ of heat is given off?
4.
For the system, is the work term + or –?
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Finding ∆Hrxn from a Measured Heat Transfer
Given a measured heat transfer for a reaction (qsurr) we can determine the ∆Hrxn.
Experiment shows 1.14 kJ of heat is needed to decompose 1.50 g of NaHCO3(s).
2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g)
∆Hrxn = ?
From the data, find ∆Hrxn.
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Hess’s Law of Heat Summation
Hess’s Law states that if a reaction is the sum of two or more other reactions, then the net ∆Hrxn is
the sum of the individual ∆Hrxn values.
A —> B + C
B —> D
A —> C + D
∆H1
∆H2
∆H1 + ∆H2 = ∆Hrxn
In other words, the overall change in enthalpy (∆H) for a process is independent of the path by
which the change occurs. Why is this true?
Consider the following reactions:
1. 2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g);
∆H1 = 128 kJ
2. Na2CO3(s) —> Na2O(s) + CO2(g);
∆H2 = 138 kJ
3. 2 NaHCO3(s) —> Na2O(s) + H2O(g) + 2 CO2(g);
∆H3 = 266 kJ = ∆H1 + ∆H2
Reaction (3) is the sum of reactions (1) and (2). If reaction 1 is carried out, followed by (2), the overall enthalpy
change will be equal to the enthalpy change for reaction (3) alone.
Bottom-line: For any reaction that can be written as the sum of simpler reactions, ∆Hrxn is the sum of the heats from
the individual reactions.
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Using Hess’ Law
Use Hess’ Law to find the unknown ∆Hrxn.
1. 2 Al(s) + 3/2 O2(g) —> Al2O3(s)
∆Hrxn = –1669.8 kJ
2. Al(s) + 3/2 Cl2(g) —> AlCl3(s)
∆Hrxn = –705.6 kJ
3. H2(g) + 1/2 O2(g) —> H2O(g)
∆Hrxn = –241.82 kJ
4. 1/2 N2(g) + 2 H2(g) + 1/2 Cl2(g) + 2 O2(g)—> NH4ClO4(s)
∆Hrxn = –295.3 kJ
5. 6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s)
∆Hrxn = ?
1. Work with only the reactants and products given in the target reaction. (All other species should cancel when summed.)
2. Reverse equations where reactants or products are on the incorrect side. Change the sign of ∆H.
3. Get the correct stoichiometric amounts (correct coefficients) of reactants and products on each side by multiplying each equation by the
appropriate factor. Multiply ∆H by the same factor. (Sometimes fractions are needed.)
4. Add the reactions together, canceling species with the same formula and phase that are on opposite sides of the reaction.
5. Check that you have obtained the target reaction.
6. Sum the ∆H’s.
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Standard Enthalpies of Formation, ∆H°f
Standard Enthalpy of Formation (∆Hf°), or Heat of Formation.
1. The enthalpy change for the formation of one mole of a compound (the product) directly from its component
elements (the reactants) in their standard states.
2.
3.
The standard state of an element is its most stable form in the physical state it exists at a pressure of 1bar
(about 1atm) and at a specified temperature.
Data for the heat of formation of almost all compounds has been measured and/or calculated under Standard
Thermochemical Conditions.
3.1. Standard State Thermochemical Conditions:
1 bar pressure ≈ 1 atm
25° C = 298.15 K
Aqueous Solutions: 1M
Pure elements: most stable form at 25° C and 1 bar.
Example: C(s) is more stable as C(s,graphite) not C(s,diamond).
C(s,graphite) is used for a formation reaction involving carbon.
Examples:
H2(g) + 1/2 O2(g) → H2O(l)
∆Hf˚ = –285.8 kJ/mol
H2(g) + 1/2 O2(g) → H2O(g)
∆Hf˚ = –241.8 kJ/mol
2 C(graphite) + 3 H2(g) + 1/2 O2(g) → CH3CH2OH(l)
∆Hf˚ = –277.7 kJ/mol
Write a standard enthalpy of formation reaction for Na2CO3(s).
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Notes on Standard Enthalpies of Formation, ∆H°f
1. The ∆H°f for all elements in their standard states are zero. Elements are not formed in nature, they already exist.
2. Most ∆H°f values are negative indicating most compounds are more “stable” than their elements.
3. ∆H°f can be used to compare stabilities of compounds. Generally, the more negative ∆H°f the more “stable”.
4. For aqueous compounds, the values refer to 1 M solutions being formed. This includes any energy changes that
occur during hydration of ions.
5. For some ionic compounds dissolved in water, you may only be able to find ∆Hf˚ values for the separate aqueous
ions. For example, for FeCl3(aq) you may not find the ∆Hf˚ value for FeCl3(aq), but may find the ions separately:
Fe3+(aq) and Cl–(aq). (In other words, you may need to use net ionic equations in some cases when you use
∆Hf˚ values.)
Why is H2O(s) not listed?
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Hess’ Law and ∆H°rxn
We can use tabulated values of ∆H°f to calculate the standard heat of reaction, ∆H°rxn, for any possible reaction.
Use Appendix C in the textbook to find ∆H°f values.
We apply Hess’ Law:
o
∆ H rxn
= ∑ n ∆ H of ( products ) − ∑ n ∆ H of ( reactants )
Example: Find ∆H°rxn for the following:
6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s)
∆H°f (kJ/mol) –295.3
0
0
–241.82
–1669.8
–705.6 Look-up values using reference source.
∆H°f (kJ) –1771.8
∑∆H°f (kJ) .
0
–1771.8 (reactants)
0
∆H°rxn (kJ) = –10992.2 – (–1771.8) = -9220.4 kJ –2901.84
–6679.2
–10992.2 (products)
–1411.2 Mult. by the molar stoichiometric coeff.
Sum products and reactants separately
Subtract reactant sum from product sum.
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Using ∆H°f to Calculate ∆H°rxn
Use Appendix C and Hess’ Law to find ∆H°rxn for each of the following:
1.
H2O(l) —> H2O(g)
2.
Fe2O3(s) + 2 Al(s) —> 2 Fe(s) + Al2O3(s)
3.
NH4Cl(aq) + AgNO3(aq) —> AgCl(s) + NH4NO3(aq)
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Thermite Welding
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Text Problems
Problem 5.9: Imagine a container placed in a tub of water, as depicted in the following diagram:
(a) If the contents of the container are the system and heat is able to flow through the container
walls, what qualitative changes will occur in the temperatures of the system and in its
surroundings?
What is the sign of q associated with each change?
From the system’s perspective, is the process exothermic or endothermic?
(b) If neither the volume nor the pressure of the system changes during the process, how is the
change in internal energy related to the change in enthalpy?
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Text Problems
Problem 5.29: A gas is confined to a cylinder fitted with a piston and an electrical heater,
as shown in the accompanying illustration:
Suppose that current is supplied to the heater so that 100 J of energy is added. Consider
two different situations. In case (1) the piston is allowed to move as the energy is added.
In case (2) the piston is fixed so that it cannot move.
(a) In which case does the gas have the higher temperature after addition of the electrical
energy? Explain.
(b) What can you say about the values of q and w in each of these cases?
(c) What can you say about the relative values of ∆E for the system (the gas in the
cylinder) in the two cases?
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Text Problems
Problem 5.104a: Calculate the standard enthalpy of formation of gaseous
diborane (B2H6) using the following thermochemical equations:
4B(s) + 3O2(g)
—> 2B2O3(s)
2H2(g) + O2(g) —> 2H2O(l)
B2H6(g) + 3O2(g) —> B2O3(s) + 3H2O(l) ∆H° = –2509.1 kJ
∆H° = –571.7 kJ
∆H° = –2147.5 kJ
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Text Problems
Problem 5.113: Consider the following unbalanced oxidation-reduction reactions
in aqueous solution:
(a) Balance each of the reactions.
(b) By using data in Appendix C, calculate ∆H° for each of the reactions.
(c) Based on the values you obtain for which of the reactions would you expect
to be thermodynamically favorable? Which would you expect to be
unfavorable?
(d) Use the activity series to predict which of these reactions should occur. Are
these results in accord with your conclusion in part (c) of this problem?
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Text Problems
Problem 5.86: The heat of combustion of ethanol, C2H5OH(l), is −1367 kJ/mol. A
batch of Sauvignon Blanc wine contains 10.6% ethanol by mass. Assuming the
density of the wine to be 1.0 g/mL, what caloric content does the alcohol (ethanol)
in a 6-oz glass of wine (177 mL) have?
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Text Problems
Problem 5.102: Meals-Ready-to-Eat (MREs) are military meals that can by
heated on a flameless heater. The heat is produced by the following reaction:
Mg(s) + 2H2O(l) —> Mg(OH)2(s) + H2(g)
Calculate the number of grams of Mg needed for this reaction to release enough
energy to increase the temperature of 75 g of water from 21°C to 79°C.
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