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4C Focus on the Human Body: Lactase, a Biological Catalyst

4C Focus on the Human Body: Lactase, a Biological Catalyst

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172



ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



PROBLEM 6.13



The reaction of acetic acid (C2H4O2) and ethanol (C2H6O) to form ethyl acetate (C4H8O2) and

water occurs only when a small amount of sulfuric acid (H2SO4) is added. Is H2SO4 a catalyst

for this reaction? What effect does H2SO4 have on the relative energies of the reactants and

products?



6.4D



FOCUS ON THE ENVIRONMENT

CATALYTIC CONVERTERS



The combustion of gasoline with oxygen provides a great deal of energy, much like the oxidation

reactions of methane and propane discussed in Section 6.2B, and this energy is used to power

vehicles. As the number of automobiles increased in the twentieth century, the air pollution they

were responsible for became a major problem, especially in congested urban areas.







FIGURE 6.3



Lactase, an Example of a Biological Catalyst



The enzyme catalyzes the

breaking of this bond.

lactose

C12H22O11

enzyme



enzyme

active site

[1]



lactase

enzyme



[2] H2O



+



+



lactase

The enzyme is the catalyst. It is

recovered unchanged in the reaction.



galactose

C6H12O6



glucose

C6H12O6



The enzyme lactase binds the carbohydrate lactose (C12H22O11) in its active site in step [1]. Lactose then reacts with water to break a

bond and form two simpler sugars, galactose and glucose, in step [2]. This process is the first step in digesting lactose, the principal

carbohydrate in milk. Without the enzyme, individuals are unable to convert lactose to galactose and glucose, lactose cannot be

metabolized, and digestive problems result.



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EQUILIBRIUM



173







HEALTH NOTE



FIGURE 6.4



How a Catalytic Converter Works

H2O

N2



catalytic converter

CO2



NO

O2



CO

gasoline



There is a direct link between the

poor air quality in large metropolitan areas like Los Angeles and an

increase in respiratory diseases.



CxHy



+



2 CO



+



+



O2



CO2



O2



2 CO2



2 NO



N2



+



H2O (unbalanced)

O2



Three reactions are catalyzed by a metal catalyst,

usually rhodium, platinum, or palladium.



A catalytic converter uses a metal catalyst—rhodium, platinum, or palladium—to catalyze three

reactions that clean up the exhaust from an auto engine.



One problem with the auto engines of the 1970s centered on the carbon- and nitrogen-containing

by-products emitted in engine exhaust. In addition to CO2 and H2O formed during combustion,

auto exhaust also contained unreacted gasoline molecules (general formula CxHy), the toxic gas

carbon monoxide (CO, Section 5.5), and nitrogen monoxide (NO, Section 5.5, a contributing

component of acid rain). Catalytic converters were devised to clean up these polluting automobile emissions.

The newest catalytic converters, called three-way catalytic converters, use a metal as a surface

to catalyze three reactions, as shown in Figure 6.4. Both the unreacted gasoline molecules and

carbon monoxide (CO) are oxidized to CO2 and H2O. Nitrogen monoxide is also converted to

oxygen and nitrogen. In this way, three molecules that contribute to unhealthy smog levels are

removed, and the only materials in the engine exhaust are CO2, H2O, N2, and O2.



PROBLEM 6.14



Nitrogen dioxide, NO2, also an undesired product formed during combustion, is converted to

N2 and O2 in a catalytic converter. Write a balanced equation for this reaction.



6.5 EQUILIBRIUM

Thus far in discussing reactions we have assumed that the reactants are completely converted

to products. A reaction of this sort is said to go to completion. Sometimes, however, a reaction

is reversible; that is, reactants can come together and form products, and products can come

together to re-form reactants.



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174



ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



• A reversible reaction can occur in either direction, from reactants to products or from

products to reactants.



Consider the reversible reaction of carbon monoxide (CO) with water to form carbon dioxide

(CO2) and hydrogen. Two full-headed arrows (

) are used to show that the reaction can

proceed from left to right and right to left as written.

The forward reaction proceeds to the right.

CO(g)



+



H2O(g)



CO2(g) +



H2(g)



The reverse reaction proceeds to the left.



• The forward reaction proceeds from left to right as drawn.

• The reverse reaction proceeds from right to left as drawn.



When CO and H2O are mixed together they react to form CO2 and H2 by the forward reaction.

Once CO2 and H2 are formed, they can react together to form CO and H2O by the reverse reaction. The rate of the forward reaction is rapid at first, and then decreases as the concentration of

reactants decreases. The rate of the reverse reaction is slow at first, but speeds up as the concentration of the products increases.

• When the rate of the forward reaction equals the rate of the reverse reaction, the net

concentrations of all species do not change and the system is at equilibrium.



The forward and reverse reactions do not stop once equilibrium has been reached. The reactants

and products continue to react. Since the rates of the forward and reverse reactions are equal,

however, the net concentrations of all reactants and products do not change.



PROBLEM 6.15



Identify the forward and reverse reactions in each of the following reversible reactions.

a. 2 SO2(g) + O2(g)

b. N2(g) + O2(g)

c. C2H4O2 + CH4O



2 SO3(g)

2 NO(g)

C3H6O2 + H2O



6.5A THE EQUILIBRIUM CONSTANT

Because the net concentrations of the reactants and products do not change at equilibrium, they

are used to define an expression, the equilibrium constant, K, which has a characteristic value

for a reaction at a given temperature. When discussing equilibrium it is not the absolute number

of moles that is the important quantity, but rather it is the concentration, the number of moles

in a given volume. Brackets, [ ], are used to symbolize concentration, which is reported in moles

per liter (mol/L).

Consider the following general reaction, where A and B represent reactants, C and D represent

products, and a, b, c, and d represent the coefficients in the balanced chemical equation.

aA



+



bB



cC



+



dD



The equilibrium constant K is the ratio of the concentrations of the products (C and D) multiplied

together, to the concentrations of the reactants (A and B) multiplied together. Each concentration

term is raised to a power equal to the coefficient in the balanced chemical equation.



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EQUILIBRIUM



175



concentration of each product (mol/L)



Equilibrium constant



=



K



=



[products]

[reactants]

concentration of each reactant (mol/L)



K



=



[C]c[D]d

[A]a[B]b



The expression for the equilibrium constant for any reaction can be written from a balanced equation, as shown for the reaction of N2 and O2 to form NO.

above the division line

Balanced equation:



N2(g)



+



O2(g)



2 NO(g)

The coefficient becomes the exponent.



below the division line

Equilibrium constant



SAMPLE PROBLEM 6.8



=



K



=



[NO]2

[N2][O2]



Write the expression for the equilibrium constant for the following balanced equation.

2 CO(g) + O2(g)



2 CO2(g)



ANALYSIS



To write an expression for the equilibrium constant, multiply the concentration of the products

together and divide this number by the product of the concentrations of the reactants. Each

concentration term must be raised to a power equal to the coefficient in the balanced chemical

equation.



SOLUTION



The concentration of the sole product, CO2, is placed in the numerator and raised to the second

power since this term has the coefficient “2.” The denominator contains concentration terms for

the two reactants, CO and O2, multiplied together. Since the coefficient preceding CO in the

balanced equation is “2,” this concentration term has an exponent of “2.”

Equilibrium constant = K =



PROBLEM 6.16



[CO2]2

[CO]2[O2]



Write the expression for the equilibrium constant for each equation.

a.

b.

c.

d.



PCl3(g) + Cl2(g)

2 SO2(g) + O2(g)

H2(g) + Br2(g)

CH4(g) + 3 Cl2(g)



PCl5(g)

2 SO3(g)

2 HBr(g)

CHCl3(g) + 3 HCl(g)



Since K is a constant, it determines the ratio of products to reactants at equilibrium no matter how

much of each substance is present at the beginning of a reaction, as shown in Figure 6.5.



6.5B THE MAGNITUDE OF THE EQUILIBRIUM CONSTANT

The magnitude of the equilibrium constant tells us whether the products or reactants are favored

once equilibrium is reached.

• When the equilibrium constant is much greater than one (K > 1), the concentration of the

products is larger than the concentration of the reactants. We say equilibrium lies to the

right and favors the products.



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176







ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



FIGURE 6.5



Equilibrium Concentrations from Different Amounts of Substances



Balanced equation:



CO(g)



+



H2O(g)



CO2(g)



Only compounds to the left of the

arrows begin the reaction.



+



H2(g)



At equilibrium



Reaction [1]

Reaction occurs.



Initial amounts per unit volume:

• 5 CO molecules

• 5 H2O molecules



Equilibrium amounts per unit volume:

• 4 CO2 molecules

• 1 CO molecule

• 4 H2 molecules

• 1 H2O molecule



Only compounds to the right of the

arrows begin the reaction.



Reaction [2]



At equilibrium



Equilibrium concentrations

are the same.



Reaction occurs.



Initial amounts per unit volume:

• 5 CO2 molecules

• 5 H2 molecules



Equilibrium amounts per unit volume:

• 1 CO molecule

• 4 CO2 molecules

• 4 H2 molecules

• 1 H2O molecule



• Reaction [1] begins with only CO and H2O molecules in equal amounts. At equilibrium there are four times as many product

molecules (CO2 and H2) as reactant molecules (CO and H2O).

• Reaction [2] begins with only CO2 and H2 molecules in equal amounts. At equilibrium, there are four times as many product

molecules (CO2 and H2) as reactant molecules (CO and H2O).

• Thus, it does not matter if the initial reaction mixture contains only compounds to the left of the equilibrium arrows or only

compounds to the right of the equilibrium arrows. The equilibrium concentrations of the products and reactants are the same.



[products]

When K is greater than 1:

[reactants]

(K > 1)



The numerator is larger.

Equilibrium favors the products.



• When the equilibrium constant is much less than one (K < 1), the concentration of the

reactants is larger than the concentration of the products. We say equilibrium lies to the

left and favors the reactants.



When K is less than 1:

(K < 1)



[products]

[reactants]



The denominator is larger.

Equilibrium favors the reactants.



• When the equilibrium constant is around 1, anywhere in the range of 0.01–100, both

reactants and products are present at equilibrium.



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EQUILIBRIUM



177



When K is approximately equal to 1:

(K 1)



[products]

[reactants]



Both reactants and products are present.



For example, the equilibrium constant for the reaction of H2 and O2 to form water is much greater

than one, so the product, H2O, is highly favored at equilibrium. A reaction with such a large K

essentially goes to completion, with little or no reactants left.

2 H2(g)



+



O2(g)



2 H2O(g)



K



=



2.9 ì 1082



The product is highly favored since K > 1.

• Equilibrium proceeds to the right.



In contrast, the equilibrium constant for the conversion of O2 to O3 is much less than one, so the

reactant, O2, is highly favored at equilibrium, and almost no product, O3, is formed. The relationship

between the equilibrium constant and the direction of equilibrium is summarized in Table 6.4.

3 O2(g)



2 O3(g)



K



=



2.7 × 10−29



• The reactant is highly favored since K < 1.

• Equilibrium proceeds to the left.



Generally there is a relationship between the equilibrium constant K and the ∆H of a reaction.

• The products of a reaction are favored when K is much greater than one (K > 1), and

∆H is negative. In other words, equilibrium favors the products when they are lower in

energy than the reactants.



There is, however, no relationship between K and the reaction rate. Some reactions with very

large equilibrium constants are still very slow. Moreover, a catalyst may speed up a reaction, but

it does not affect the size of K. With a catalyst, equilibrium is reached more quickly, but the relative concentrations of reactants and products do not change.



TABLE 6.4



How the Magnitude of K Relates to the Direction

of Equilibrium



Value of K



PROBLEM 6.17



K>1



Equilibrium favors the products.

Equilibrium lies to the right.



K<1



Equilibrium favors the reactants.

Equilibrium lies to the left.



K≈1



Both the reactants and products are present at equilibrium.



Given each equilibrium constant, state whether the reactants are favored, the products are

favored, or both the reactants and products are present at equilibrium.

a. 5.0 × 10–4



smi26573_ch06.indd 177



Position of Equilibrium



b. 4.4 × 105



c. 35



d. 0.35



PROBLEM 6.18



Predict whether the reactions with the equilibrium constants in Problem 6.17 are endothermic

or exothermic.



PROBLEM 6.19



Consider the following reaction: 2 H2(g) + S2(g)

2 H2S(g) with K = 1.1 × 107. (a) Write

the expression for the equilibrium constant for this reaction. (b) Are the reactants or products

favored at equilibrium? (c) Would you predict ∆H to be positive or negative? (d) Are the

reactants or products lower in energy? (e) Would you predict this reaction to be fast or slow?

Explain your choice.



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178



ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



6.5C



CALCULATING THE EQUILIBRIUM CONSTANT



The equilibrium constant for a reaction is an experimentally determined value, and thus it can be

calculated if the concentrations of all substances involved in a reaction are measured at equilibrium. Concentrations are reported in moles per liter (mol/L), symbolized as M.



HOW TO

EXAMPLE



Calculate the Equilibrium Constant for a Reaction

Calculate K for the reaction between the general reactants A2 and B2. The equilibrium concentrations are as

follows: [A2] = 0.25 M; [B2] = 0.25 M; [AB] = 0.50 M.

A2 + B2



Step [1]



2 AB



Write the expression for the equilibrium constant from the balanced chemical equation, using the coefficients as

exponents for the concentration terms.

Equilibrium constant = K =



Step [2]



[AB]2

[A2][B2]



Substitute the given concentrations in the equilibrium expression and calculate K.

• Since the concentration is always reported in mol/L (M), these units are omitted during the calculation.

K =



[0.50]2

[AB]2

(0.50) × (0.50)

=

=

[A2][B2]

[0.25][0.25]

0.0625

0.25

= 0.0625 = 4.0

Answer



SAMPLE PROBLEM 6.9



Calculate K for the reaction of N2 and H2 to form NH3, with the given balanced equation and

the following equilibrium concentrations: [N2] = 0.12 M; [H2] = 0.36 M; [NH3] = 1.1 M.

N2(g) + 3 H2(g)



ANALYSIS



2 NH3(g)



Write an expression for K using the balanced equation and substitute the equilibrium

concentrations of all substances in the expression.



SOLUTION

K =



[NH3]2

3



[N2][H2]



=



(1.1)2

(0.12)(0.36)3



=



=



1.1 × 1.1

0.12 × 0.36 × 0.36 × 0.36

1.21

0.0056



= 216 rounded to 220

Answer



PROBLEM 6.20



Calculate the equilibrium constant for each reaction using the balanced chemical equations and

the concentrations of the substances at equilibrium.

a. CO(g) + H2O(g)

b. 2 NO2(g)



smi26573_ch06.indd 178



CO2(g) + H2(g)

N2O4(g)



[CO] = 0.0236 M; [H2O] = 0.00240 M;

[CO2] = 0.0164 M; [H2] = 0.0164 M

[NO2] = 0.0760 M; [N2O4] = 1.26 M



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LE CHÂTELIER’S PRINCIPLE



179



6.6 LE CHÂTELIER’S PRINCIPLE

What happens when a reaction is at equilibrium and something changes? For example, what

happens when the temperature is increased or some additional reactant is added? Le Châtelier’s

principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium. Le Châtelier’s principle states:

• If a chemical system at equilibrium is disturbed or stressed, the system will react in the

direction that counteracts the disturbance or relieves the stress.



Let’s examine the effect of changes in concentration, temperature, and pressure on equilibrium.



6.6A CONCENTRATION CHANGES

Consider the reaction of carbon monoxide (CO) with oxygen (O2) to form carbon dioxide (CO2).

2 CO(g) + O2(g)



The conversion of nitrogen and

hydrogen to ammonia (Sample

Problem 6.9) is an exceedingly

important reaction since it converts

molecular nitrogen from the air to a

nitrogen compound (NH3) that can

be used as a fertilizer for plants. In

the early twentieth century, German

chemist Fritz Haber developed a

catalyst that enabled this reaction to

proceed quickly to afford favorable

yields of NH3. This process paved

the way for large-scale agriculture,

which provides food for the world’s

growing population.



2 CO2(g)



If the reactants and products are at equilibrium, what happens if the concentration of CO is

increased? Now the equilibrium is disturbed and, as a result, the rate of the forward reaction

increases to produce more CO2. We can think of added reactant as driving the equilibrium

to the right.

Adding more reactant...



2 CO(g)



+



O2(g)



2 CO2(g)



...drives the reaction to the right.



When the system reaches equilibrium once again, the concentrations of CO2 and CO are both

higher. Since O2 reacted with the additional CO, its new value at equilibrium is lower. Even

though the concentrations of reactants and products are different, the value of K is the same.

If CO is added...

2 CO(g)



+



O2(g)



2 CO2(g)



...this decreases...



...and this increases.



What happens, instead, if the concentration of CO2 is increased? Now the equilibrium is disturbed but there is more product than there should be. As a result, the rate of the reverse reaction

increases to produce more of both reactants, CO and O2. We can think of added product as

driving the equilibrium to the left.

Adding more product...



2 CO(g)



+



O2(g)



2 CO2(g)



...drives the reaction to the left.



When the system reaches equilibrium once again, the concentrations of CO, O2, and CO2 are all

higher. Even though the concentrations of reactants and products are different, the value of K is

the same.



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180



ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



If CO2 is added...

2 CO(g)



+



...this increases...



O2(g)



2 CO2(g)



...and this increases.



Similar arguments can be made about the effect of decreasing the concentration of a reactant or

product. Sometimes, when K < 1 and the amount of product at equilibrium is not high, a product

is removed from the reaction mixture as it is formed. For example, ethanol (C2H6O) can be converted to ethylene (CH2 CH2) and water in the presence of a small amount of acid, but equilibrium does not favor the products.

Removing a product...



H

C2H6O

ethanol



H

C



+



C



H



H2O



H



...drives the reaction to the right.



In this case, water can be removed from the reaction as it is formed. A decrease in concentration

of one product results in more of the forward reaction to form more product. This process drives

the equilibrium to the right. If water is continuously removed, essentially all of the starting

material can be converted to product.



SAMPLE PROBLEM 6.10



In which direction is the equilibrium shifted with each of the following concentration changes for

the given reaction: (a) increase [SO2]; (b) increase [SO3]; (c) decrease [O2]; (d) decrease [SO3]?

2 SO2(g) + O2(g)



ANALYSIS



SOLUTION



PROBLEM 6.21



2 SO3(g)



Use Le Châtelier’s principle to predict the effect of a change in concentration on equilibrium.

Adding more reactant or removing product drives the equilibrium to the right. Adding more

product or removing reactant drives the equilibrium to the left.

a.

b.

c.

d.



Increasing [SO2], a reactant, drives the equilibrium to the right to form more product.

Increasing [SO3], a product, drives the equilibrium to the left to form more reactants.

Decreasing [O2], a reactant, drives the equilibrium to the left to form more reactants.

Decreasing [SO3], a product, drives the equilibrium to the right to form more product.



In which direction is the equilibrium shifted with each of the following concentration changes for

the given reaction: (a) increase [H2]; (b) increase [HCl]; (c) decrease [Cl2]; (d) decrease [HCl]?

H2(g) + Cl2(g)



6.6B



2 HCl(g)



TEMPERATURE CHANGES



In order to predict what effect a change of temperature has on a reaction, we must know if a reaction is endothermic or exothermic.

• When temperature is increased, the reaction that removes heat is favored.

• When temperature is decreased, the reaction that adds heat is favored.



For example, the reaction of N2 and O2 to form NO is an endothermic reaction (∆H = +43 kcal/mol).

Since an endothermic reaction absorbs heat, increasing the temperature increases the rate of

the forward reaction to form more product. The equilibrium shifts to the right.



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LE CHÂTELIER’S PRINCIPLE



181



Increasing temperature...



N2(g)



+



O2(g)



2 NO(g)



This reaction

absorbs heat.



∆H = +43 kcal/mol

endothermic



...drives the reaction to the right.



In the exothermic reaction of N2 and H2 to form NH3 (∆H = –22 kcal/mol), increasing the temperature increases the rate of the reverse reaction to form more reactants. When temperature

is increased, the reaction that absorbs heat, in this case the reverse reaction, predominates and the

equilibrium shifts to the left.

Increasing temperature...



N2(g)



+



3 H2(g)



2 NH3(g)



...drives the reaction to the left.



SAMPLE PROBLEM 6.11



∆H = −22 kcal/mol

exothermic

The reverse reaction

absorbs heat.



The reaction of SO2 with O2 to form SO3 is an exothermic reaction. In which direction is the

equilibrium shifted when the temperature is (a) increased; (b) decreased?



ANALYSIS



When temperature is increased, the reaction that removes heat is favored. When temperature is

decreased, the reaction that adds heat is favored.



SOLUTION



In an exothermic reaction, the forward reaction releases heat and the reverse reaction absorbs

heat. (a) When the temperature is increased, the reverse reaction is favored because it removes

heat, so the equilibrium shifts to the reactants (to the left). (b) When the temperature is

decreased, the forward reaction is favored because it adds heat, so the equilibrium shifts to the

products (to the right).



PROBLEM 6.22



The conversion of H2O to H2 and O2 is an endothermic reaction. In which direction is the

equilibrium shifted when the temperature is (a) increased; (b) decreased?



PROBLEM 6.23



The reaction of O3 with NO to form NO2 and O2 is an exothermic reaction. In which direction

is the equilibrium shifted when the temperature is (a) increased; (b) decreased?



6.6C



PRESSURE CHANGES



When the substances involved in a reaction are gases and the number of moles of reactants and

products differs, a change in pressure has an effect on equilibrium.

Pressure, the force per unit area,

is discussed in greater detail in

Chapter 7.



• When pressure increases, equilibrium shifts in the direction that decreases the number

of moles in order to decrease pressure.

• When pressure decreases, equilibrium shifts in the direction that increases the number

of moles in order to increase pressure.



In the reaction of N2 and H2 to form NH3, there are four moles of reactants but only two moles

of product. When the pressure of the system is increased, the equilibrium shifts to the right since

there are fewer moles of product.



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182



ENERGY CHANGES, REACTION RATES, AND EQUILIBRIUM



Increasing pressure...



N2(g)



+



3 H2(g)



2 NH3(g)



4 moles



2 moles

This side has fewer moles.



...drives the reaction to the right.



In contrast, when the pressure of the system is decreased, the equilibrium shifts to the left since

there are more moles of reactants.

Decreasing pressure...



N2(g)



+



3 H2(g)



2 NH3(g)



4 moles



This side has

more moles.



2 moles



...drives the reaction to the left.



SAMPLE PROBLEM 6.12



In which direction is the equilibrium shifted in the following reaction when the pressure is

(a) increased; (b) decreased?

H



C



C



H(g) + 2 H2(g)



C2H6(g)



ANALYSIS



When pressure is increased, the equilibrium shifts in the direction that decreases the number

of moles. When pressure is decreased, the equilibrium shifts in the direction that increases the

number of moles.



SOLUTION



In this reaction, there are three moles of reactants and only one mole of product. (a) Increasing

the pressure shifts the equilibrium to the right, to the side of fewer moles. (b) Decreasing the

pressure shifts the equilibrium to the left, to the side of more moles.



PROBLEM 6.24



In which direction is the equilibrium shifted in the following reaction when the pressure is

(a) increased; (b) decreased?

C2H4(g) + Cl2(g)



C2H4Cl2(g)



Table 6.5 summarizes the effects of changes in reaction conditions on the direction of an equilibrium.



6.7 FOCUS ON THE HUMAN BODY

BODY TEMPERATURE

The human body is an enormously complex organism that illustrates important features of energy

and reaction rates. At any moment, millions of reactions occur in the body, when nutrients are

metabolized and new cell materials are synthesized.

Normal body temperature, 37 °C, reflects a delicate balance between the amount of heat absorbed

and released in all of the reactions and other processes. Since reaction rate increases with increasing temperature, it is crucial to maintain the right temperature for proper body function. When



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