reaction energy and kinetics - ms. tabors classroom · web viewis the si unit for energy. one joule...

Name: ______________________Chemistry 1 Notes, 2012–13; P. Holloman

Topic 9 – Energetics & Kinetics

2.1.1 – Explain the energetic nature of phase changes.2.1.2 – Explain heating and cooling curves (heat of fusion, heat of vaporization, heat, melting point, and boiling point.2.1.4 – Infer simple calorimetric calculations based on the concepts of heat lost equals heat gained and specific heat.2.2.1 – Explain the energy content of a chemical reaction.3.1.1 – Explain the factors that affect the rate of a reaction (temperature, concentration, particle size and presence of a catalyst)

I. Energy♦ All energy falls under two headings: kinetic and potential.

1. kinetic energy – energy of motion (falling rock, moving atoms)2. potential energy – energy that an object has because of its position

or composition (water behind a dam, gasoline in a car)

Law of Conservation of Energy: Energy can be converted from one form to another (e.g. potential to kinetic), but it cannot be created or destroyed in ordinary chemical or physical changes. Another way to think of this law is to say that, for a closed system, energy

is neither lost nor gained in a normal chemical reaction.

TEMPERATURE VS. HEAT

Let’s also recall the difference between heat and temperature from our first unit:

Temperature is the average kinetic energy of the particles in a substance. Temperature is usually measured in units of Celsius or Kelvin.

Heat is the total kinetic energy of the particles in a substance. Heat is usually measured in units of joules or kilojoules. The joule (J) is the SI unit for energy. One joule is the work done, or energy expended, by a

force of one Newton moving an object one meter along the direction of the force. This quantity is also denoted as a Newton-meter with the symbol N·m.

1 J =

1 kg m2

s2 = 1 N·m

In case you need to convert: 1 kcal = 1 Cal = 4.184 kJ

To distinguish them, imagine a bathtub filled with room temperature water and a cup filled with boiling water. The water in the cup has a higher temperature (on average, the water molecules in the cup are moving faster). But the bathtub has more heat (because there’s much more water).

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ENERGY IN REACTIONS

Enthalpy (H) is the heat content of a system. H = change in enthalpy; the amount of heat released or absorbed during a reaction.

The heat of reaction (enthalpy of reaction) is the exact amount of heat energy (usually in kilojoules) released or absorbed during a chemical reaction. Heat of reaction is often expressed as a change in enthalpy (H). H is negative for exothermic reactions, positive for endothermic reactions

A thermochemical equation shows the amount of heat absorbed or released during that reaction.

In exothermic reactions, heat is released (as a product) in the reaction2H2(g) + O2(g) 2H2O(g) + 483.6 kJ (H = – 483.6 kJ)

endothermic rxns. – heat must be absorbed for the reaction to go (as a reactant)N2(g) + O2(g) + 90.3 kJ 2NO(g) (H = + 90.3 kJ)

II. The Reaction Process In order for a reaction to occur, the reactant molecules must collide with each other in just the right

way. The colliding molecules must possess:

1. the correct collision geometry (orientation)2. sufficient kinetic energy (activation energy)

to form the activated complex…

activated complex – an unstable, high-energy, transitional structure, resulting from a collision between reactant molecules, which exists while old bonds are breaking and new ones are forming

activation energy – energy required to transform the reactants into the activated complex

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(Activated Complex)

When reactants collide, one of two things may happen:

1) They may form the activated complex, then go on and form new products, if they possess the correct collision geometry / orientation (they are turned the right way so that the bonds in the activated complex can form) and sufficient activation energy (they are moving fast enough when they collide that they can form the unstable high-energy bonds of the activated complex).

2) They may bounce off of each other and remain the original reactants, either because they collide too slowly (not enough activation energy) or they are not turned the right way to form the activated complex bonds (incorrect orientation / collision geometry).

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REACTION PATHWAYS

EXOTHERMIC REACTIONS

Consider what an exothermic reaction would look like if you plotted the potential energy of all the species involved vs. time (or reaction progress). Obviously, the activated complex is the least stable species involved, and thus has the highest

potential energy. The reactants must collide with enough energy to form the high-energy activated complex.

In an exothermic reaction, heat is given off because the products are more stable than the reactants (the products possess a lower potential energy). The excess bond energy not needed by the products is radiated off in the form of heat.

Consider the PE diagrams below for glycolysis, and for a generalized exothermic reaction:

ENDOTHERMIC REACTIONS

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In this exothermic reaction: EA = +10 kJ (forward reaction) EA’ = +30 kJ (reverse reaction) Hforward = –20 kJ Hreverse = +20 kJ

This figure could represent an exothermic reaction. (A) represents the potential energy of the reactants, which must be pushed up a hill. The top of the hill (where the boulder is least stable) is like the activated complex. Once you get past the top of the hill, the reaction proceeds spontaneously until it becomes stable again (at the bottom of the hill) – when the products have been formed.


PE diagrams for an endothermic reaction can be illustrated generally for any reaction as:

Here are potential energy diagrams for two different decomposition reactions. On the left, the decomposition of ozone (with and without a chlorine catalyst). On the right, the decomposition of KClO3 into KCl and O2 (with and without a catalyst).

Note that catalysts act by lowering the activation energy required for a reaction (often by splitting the reaction up into multiple steps, which have lower activation energies. This speeds up a reaction because when reactant molecules collide, they are more likely to have sufficient activation energy to react (since less is required). Economically speaking, this saves companies money, because they would otherwise have to heat up the reaction to achieve adequate numbers of particles with sufficient activation energy (which means higher electricity costs and perhaps special equipment).

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In this endothermic PE Diagram: A = EA for the forward reaction B = PE of the activated complex C = H D = EA’ for the reverse reaction


Whether endothermic or exothermic, it’s like the reactants lie in an energy trough – they have to build up enough momentum (kinetic energy) so that when they collide, they have sufficient activation energy to form the activated complex (which has the highest potential energy of anything on the PE diagram, being more unstable than either the reactants or products).

This is true even for an exothermic reaction! It’s sort of an endothermic component at the beginning of a reaction that ends up releasing more energy (in forming products) than it had to absorb to form the activated complex. So exothermic reactions can proceed spontaneously on their own, but only given sufficient activation energy (from the ambient heat in their surroundings) to form the activated complex.

A practical example of this is the spark plug in your car’s engine. Your car burns gasoline for energy. Gasoline is a mixture of hydrocarbons (especially octane

and its isomers) that are combusted in the engine to move the pistons up and down, which ultimately turn the axle and the wheels on the car.

Combustion reactions are good examples of exothermic reactions:

2C8H18(g) + 25O2(g) ⃗ 16CO2(g) + 18H2O(g) + 10,966 kJ

The products of all combustion reactions (CO2 + H2O) are much more stable than the reactants (a hydrocarbon and O2), hence a great deal of heat energy (excess bond energy) is given off. The same is true for the combustion of all hydrocarbons.

But octane doesn’t just spontaneously burst into flames at room T. The small spark from the spark plug provides the activation energy necessary to initiate this combustion reaction.

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III. Reaction Rate The rate of a chemical reaction can be thought of as a change in concentration of a reactant or

product over time (or, in the case of a gas, as a change in pressure over time). For example – as a reaction proceeds, the concentration of the reactants will decrease, as the

concentration of the products increases.

Reaction rate is obviously proportional to the number of effective collisions between reactant molecules. More collisions = more opportunities for a reaction to occur.

Factors affecting reaction rate :

1) The Reactants (the nature of the reactants) By “nature of the reactants”, we can mean things like the number and type of bonds that must

be broken and reformed in the course of a reaction (and the general structural complexity of the reactant and product molecules).

H2 combines with F2, but barely reacts with N2 at all under the exact same conditions.

2) Surface Area More surface area = greater # of collisions possible per unit time = faster reaction.

3) Concentration Increasing the concentration of a reactant usually increases the rate of the reaction. More particles in the same amount of space are more likely to bump into each other. ex: If you place burning charcoal in pure O2, it will burn much faster than in air,

which is only O2. Consider this idea symbolized in the diagrams

here. Doubling the concentration of the red reactant doubles the frequency of collisions with the blue reactant molecules, thus doubling the rate of reaction.

3b) Pressure Pressure is only a factor on reaction rate for reactions involving gases. Changing the pressure

on a reaction involving only solids or liquids has no effect on the rate. Increasing pressure on a reaction where the reactants are gases will increase the rate of the

reaction. Increasing pressure on a gas is exactly the same as increasing its concentration. When

you increase the pressure, you squeeze the gas molecules together into a smaller volume. (In this discussion, we’re assuming you’re holding the temperature constant.) The same number of gas particles packed together in a smaller volume means an increase in concentration, and thus a greater frequency of collisions between reactant molecules. (Refer to concentration, above…)

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4) Temperature Increase in T increase in KE increase in reaction rate. Why? Because more KE (1) greater frequency of collisions & (2) more particles with

sufficient activation energy to react when collisions occur.

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Presence of Catalysts A catalyst is a substance that changes the rate of a reaction without itself being changed. As

we previously mentioned, most catalysts speed up reactions by lowering the activation energy required for a reaction (often by splitting the reaction up into multiple steps, which have lower activation energies. This speeds up a reaction because when reactant molecules collide at any given temperature, they are more likely to have sufficient activation energy to react (since less is required).

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IV. Calorimetry Calorimetry is the study of heat flow and heat measurement. Calorimetry experiments determine heats of reaction (enthalpy changes,

or H) by measuring temperature changes produced in a calorimeter.

HEAT CAPACITY

An exothermic reaction releases heat to its surroundings, and the temperature of the surroundings increases.

The size of the T increase depends on: how much heat is released the heat capacity of the surroundings

The heat capacity of an object is the amount of heat needed to raise its T by 1C. Ex: The heat capacity of a cup of water at 18C is the # of joules of energy needed to raise

the temperature of the water to 19C. The heat capacity of an object depends on its mass and composition.

In terms of mass, the more of a substance you have, the greater its ability to absorb heat without a significant temperature change (= greater heat capacity). The flame from a match can heat a drop of water to boiling, but barely warm a full cup of water.

In terms of composition, say you have 100 mL of water and 100 mL of cyclohexane in front of you. You heat both of them equally on a hot plate. The temperature of the cyclohexane will increase much more than the water. Water, due to its nature, simply has a higher heat capacity that an equal amount of cyclohexane. (Water has a greater ability to absorb heat without a significant temperature change than most substances.)

SPECIFIC HEAT CAPACITY (SPECIFIC HEAT)

Specific heat capacity is the heat capacity of 1 gram of a substance – in other words, how much heat is required to raise the temperature of 1 gram of a substance by 1C. Specific heat allows you to compare the ability of substances to absorb heat, regardless of

mass (since you’re looking at heat capacity per gram). Specific heat is a physical property (like color and melting point). The specific heat of water

is 4.184 J g –1 C–1 (or 1.00 cal g–1 C–1). Even if you add a great deal of heat energy to a substance, you still might get only a small

temperature change, if that substance has a high specific heat.

CALORIMETERS

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A calorimeter consists of a well-insulated container filled with a known quantity of water, where a reaction takes place.

You measure the initial temperature, Ti (before the reaction) and the final temperature, Tf, to find the temperature change (T).

♦ A simple calorimeter can be constructed out of a pair of nestled polystyrene coffee cups, with a lid (to reduce heat loss). Insert a thermometer through a hole in the lid, fill the coffee cup with a known mass of water (or an aqueous solution), then conduct your experiment and record the temperature changes.

Consider how you might calculate the heat of solution (Hsoln) for dissolving 10.0g of NaOH in water. Start by setting up a coffee cup calorimeter as described above. Then measure out 100 mL of distilled water in a graduated cylinder, and pour it into your calorimeter. Measure & record the initial temperature of the water with a thermometer. Weigh out the NaOH on a digital balance, and add the NaOH to the water in the calorimeter. Quickly put the lid back on, stirring gently with the thermometer or stirrer, as shown here). Watch the thermometer and record the maximum temperature reached (NaOH dissolving in water is exothermic, so the water’s temperature will go up). Subtract Tf – Ti to find T, which you use to calculate Hsoln.

The amount of heat absorbed by the water (Hsoln) can be represented by:

q = mCpT

where:q = enthalpy change (H, in J or kJ)m = mass of water (water has a density of 1.0 g mL–1, so the volume of water (mL)

is exactly equal to its mass (in grams))Cp = specific heat of water (4.18 kJ kg–1 C–1 or 4.18 kJ kg–1 K–1)

or (4.18 J g–1 C–1 or 4.18 J g–1 K–1) (we’re dealing with T, and a of 1C = a of 1K)

T = temperature change

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You can assume that the specific heat of dilute aqueous solutions is the same as that of pure water…

A calorimeter is filled with 75.0 g of water at an initial temperature of 19.8C. A 0.050 mol sample of solid NaOH is added, and the temperature increases to 26.7C. Determine the quantity of heat (enthalpy change) for this solution process.

(ans. = +2170 J)

You can also determine heat transfer based on liquid/solution volume, instead:

q = vCpT

Here, v = volume of water (solution); and Cp = specific heat of water (or whatever the object/solvent is)

(the specific heat of water is 4.184 J mL–1 C–1 or 4.184 kJ L–1 C–1) (again, assume that the specific heat of dilute solutions is the same as water)

When a 4.25 g sample of solid ammonium nitrate (NH4NO3) dissolves in 60.0 mL of water in a calorimeter, the temperature drops from 21.0C to 16.9C. Determine the enthalpy change for this solution process. (ans. = –1.0 x 103 J)

The molar enthalpy for a reaction (or physical change, such as this dissolution) is the change in enthalpy per mole of a reactant (or product, for that matter) or solute. Once you’ve calculated the enthalpy change for the reaction / process, often using q = mCpT, as in the previous problem, it’s pretty easy. As long as you’re calculating the molar enthalpy of a reactant (and you usually are), you simply flip the sign on the enthalpy (from negative to positive or vice versa) and divide by the number of moles of the reactant in question. You have to flip the sign because the thermometer was measuring T for the water (the solution produced by the reaction / dissolution) – not the T for the reactant itself! (This would often be impossible to measure directly.) But we will assume that all the heat given off by the reactant was absorbed by the water (or vice versa), so we just flip the sign on H to find molar enthalpy. (In other words, qlost = –qgain in water…)

Calculate the molar enthalpy of solution for ammonium nitrate from the previous problem…

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Here is a second example of using a Styrofoam coffee cup calorimeter set-up to find the enthalpy of a reaction. Let’s consider an acid-base neutralization this time – specifically, the neutralization of sodium hydroxide by hydrochloric acid, by the following equation: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

50.0 cm3 of 1.00 mol dm–3 hydrochloric acid solution was added to 50.0 cm3 of 1.00 mol dm–3

sodium hydroxide solution in a polystyrene cup. The initial temperature of both solutions was 16.7C. After stirring and accounting for heat loss, the highest temperature reached was 23.5C. Calculate the enthalpy change for this reaction. Also calculate the molar enthalpy of neutralization for sodium hydroxide, based on these results.

(Hint: In calorimetry problems where you mix two solutions

together, don’t forget to add up their total volume! You still assume their specific heat capacity of these dilute aqueous solutions is the same as that of pure water. You can also assume their density is the same as that of water.)

Calorimeters can be used to determine H for both chemical changes (reactions) and physical changes (changes of state), which we will consider next…

ENTHALPY AND PHASE CHANGES

Consider the heating and cooling curve for water, shown below:

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Recall that temperature is average kinetic energy. Whenever the T increases (whether it is a solid, liquid or gas), energy has been absorbed by the

H2O, and whenever T decreases, energy is released by the H2O to the environment. Whenever the temperature of the H2O changes (regardless of phase), you can calculate the

corresponding change in enthalpy via: q = mCpT However, during phase changes, there is only a change in potential energy as intermolecular

forces are being overcome (such as H-bonding in water) and the interparticle spacing changes. Thus, with no change in KE (and therefore no T change), you cannot use q = mCpT.

Enthalpy changes (no matter which formula you use) are directly related to the number of moles in the reaction (or grams, obviously). Doubling the number of moles will double the value of H, etc. As one mole of ice melts (an endothermic phase change), it has an enthalpy change of +6.01 kJ.

So the molar enthalpy constant for melting water (the heat of fusion for water) would be:

Hmelting = Hfusion = +6.01 kJ mol–1 or +334 J g–1

The molar enthalpy constant for the reverse process (freezing) would be:

Hfreezing = –6.01 kJ mol–1 or –334 J g–1

One mole of boiling water (also an endothermic phase change), has an enthalpy change of +40.67 kJ. So the molar enthalpy constant for boiling water (the heat of vaporization for water) would be:

Hvaporization = +40.67 kJ mol–1 or +2260 J g–1

Since you can’t use q = mCpT to determine enthalpy changes during phase changes, you would have to use similar formulas – formulas that didn’t rely on temperature change. These formulas include: q = mHf (mass x heat of fusion) q = mHv (mass x heat of vaporization)

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What would be the total enthalpy change for completely boiling off 28.6 g of water at an initial temperature of 21.0C?

Look back at the heating / cooling curve for water and consider how energy and intermolecular forces are related to changes of state:

–25 – 0C: The solid particles increase in temperature as their kinetic energy increases due to greater vibration of the solid particles in a fixed position.

melting: At a certain temperature (0C for water), the vibration is sufficient to overcome the attractive forces holding the crystal lattice together, and the solid melts. During melting, the temperature does not rise as the heat energy (enthalpy of fusion) is needed to overcome these attractive forces.

0C – 100C: Once all of the solid has melted, the liquid particles move faster and the temperature increases. During the process of warming the liquid, some particles move faster than others, and have sufficient KE to escape from the surface of the liquid to form a vapour.

boiling: Once the vapour pressure is equal to atmospheric pressure (the pressure above the liquid), the liquid will boil. During boiling, the temperature remains constant as the heat is used to overcome the intermolecular forces of attraction between the liquid molecules (enthalpy of vaporization).

above 100C: When all of the liquid has turned into a gas, the temperature continues to increase as the vapor molecules move ever faster.

♦ Take a look at this excellent animation of changes of state and heating curves: http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

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http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

reaction energy and kinetics - ms. tabors classroom · web viewis the si unit for energy. one joule...

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