# An end and a new beginning…

Check out the prezi for a summary of our 2 week experiment

You’ll find out what our students thought of the experiment. I hope you enjoyed it as much as we did and find their reflections helpful!

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# Understanding the significance of Hess’s Law

TEnthalpy of formations for products and reactants (found in Appendix C) can be used via Hess’s law to calculate ∆Hrxn. These are tabulated at standard conditions (T = 25 °C and P = 1 atm). What is Hess’s Law and what do we use it for?

Germain Henri Hess (1802 – 1850) is important primarily for his thermochemical studies. The law states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps.

Hess’s law is understood as an expression of the principle of the conservation of energy, the first law of thermodynamics.

Hess’s law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the chemical equation of reactions using previously determined values for the enthalpies of formation.

Addition of chemical equations leads to a net or overall equation. If enthalpy change is known for each equation, the result will be the enthalpy change for the net equation. If the net enthalpy change is negative (ΔHnet < 0), the reaction is exothermic and is more likely to be spontaneous; positive ΔH values correspond to endothermic reactions.Entropy also plays an important role in determining spontaneity, as some reactions with a positive enthalpy change are nevertheless spontaneous.

Hess’s Law states that enthalpy changes are additive. Thus the ΔH for a single reaction:

where the o superscript indicates standard state values.

For example, in the diagram below, we look at the oxidation of carbon into CO and CO2. The direct oxidation of carbon (graphite) into CO2 yields an enthalpy of -393 kJ/mol. When carbon is oxidized into CO and then CO is oxidized to CO2, the enthalpies are -110 and -283 kJ/mol respectively. The sum of enthalpy in the two steps is exactly -393 kJ/mol, same as the one-step reaction.



The two-step reactions are:

Adding the two equations together and cancel out the intermediate, CO, on both sides leads to:

More importantly, we need to evaluate the enthalpy of CO.


Hess’s law can be applied to calculate enthalpies of reactions that are difficult to measure. In the above example, it is very difficult to control the oxidation of graphite to give pure CO. However, enthalpy for the oxidation of graphite to CO2 can easily be measured. So can the enthalpy of oxidation of CO to CO2. The application of Hess’s law enables us to estimate the enthalpy of formation of CO. Since,

C + O2-> CO2,       dH° = -393 kJ/molCO +1/2O2-> CO2,       dH° = -283 kJ/mol.

Subtracting the second equation from the first gives

C +1/2O2-> CO,             dH° = -393 -(-283) = -110 kJ/mol

The equation shows the standard enthalpy of formation of CO to be -110 kJ/mol.

Application of Hess’s law enables us to calculate dH, dH°, and dHf for chemical reactions that impossible to measure, providing that we have all the data of related reactions.

If you need more example problems, check out the following site:

http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html

ENJOY!!!

-Nnana Amakiri


References:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html

http://en.wikipedia.org/wiki/Hess's_law


# △E….LOL

Following up my last post I would like to go a step further by talking about the conservation of energy and the definition of enthalpy. What is another way to derive △E?

Another way to derive △E for the system is from LOL diagrams as done in physics. If you do not know what LOL diagrams are, please read the following blog that does a great job of explaining the concept from my former physics teacher:

http://kellyoshea.wordpress.com/2012/03/05/energy-bar-charts-lol-diagrams/

The first law of thermodynamics is the application of the conservation of energy principle to heat and thermodynamic processes. Several equations come to mind when looking at this concept:

∆H = ∆E + ∆PV (from definition) and ∆E = q + w. For example, at constant pressure, q = ∆H.

The change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by it.

Four quantities called “thermodynamic potentials” are useful in the chemical thermodynamics of reactions and non-cyclic processes. They are internal energy, the enthalpy, the Helmholtz free energy and the Gibbs free energy. Enthalpy is defined by

H = E + PV

P is your pressure, V is your volume, and E is your internal energy. The internal energy E might be thought of as the energy required to create a system in the absence of changes in temperature or volume. But if the process changes the volume, as in a chemical reaction which produces a gaseous product, then work must be done to produce the change in volume. For a constant pressure process the work you must do to produce a volume change ΔV is PΔV. Then the term PV can be interpreted as the work you must do to “create room” for the system if you presume it started at zero volume.

When work is done by a thermodynamic system, it is ususlly a gas that is doing the work. The work done by a gas at constant pressure is:

For non-constant pressure, the work can be visualized as the area under the pressure-volume curve which represents the process taking place. The more general expression for work done is:

Work done by a system decreases the internal energy of the system, as indicated in the First Law of Thermodynamics.Understanding how the conservation of energy, work, and enthalpy function we learn a new way to find △E from LOL diagrams or given data. We can manipulate the various formulas in order to solve for △E, our system’s internal energy. When we look at an lol diagram, we can find out what we need to put in the various formulas in order to manipulate them.

-Nnana Amakiri

References:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/thermo_4.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html

http://kellyoshea.wordpress.com/2012/03/05/energy-bar-charts-lol-diagrams/

# Heat in Chemistry

While working on a few chemistry problems I came across mild difficulty in finding the sign of heat. Heat is the transfer of energy that results from the difference in temperature between a system and its surroundings. At a molecular level, heat is the transfer of energy that makes use of or stimulates disorderly molecular motion in the surroundings. For instance, when a hydrocarbon fuel burns, the energy released in the reaction stimulates the surrounding atoms and molecules into more vigorous random motion, and we refer to this escape of energy as heat. Heat is not stored: Heat is energy in transit. So how do we solve for heat?

Thermochemistry is the study of the energy and heat associated with chemical reactions and/or physical transformations. A reaction may release or absorb energy, and a phase change may do the same, such as in melting and boiling. Thermochemistry focuses on these energy changes, particularly on the system‘s energy exchange with itssurroundings. Thermochemistry is useful in predicting reactant and product quantities throughout the course of a given reaction. In combination with entropy determinations, it is also used to predict whether a reaction is spontaneous or non-spontaneous, favorable or unfavorable.

Endothermic reactions absorb heat and exothermic reactions release heat. The measurement of quantities of energy transferred as heat is called calorimetry. Such a measurement is commonly made by observing the rise in temperature caused by the process being studied and interpreting that rise in terms of the heat produced. Calorimetry is used to measure the changes in internal energy and enthalpy that accompany chemical reactions.

The sign conventions for heat, work, and internal energy are summarized in the figure below. The internal energy and temperature of a system decrease (E < 0) when the system either loses heat or does work on its surroundings. Conversely, the internal energy and temperature increase (E > 0) when the system gains heat from its surroundings or when the surroundings do work on the system.

Our first law of thermodynamics says Esys=q +w

Esys stands for E system which is the total Energy for the system. Q stands for heat inputted or released and W for work done on the system.

The system is usually defined as the chemical reaction and the boundary is the container in which the reaction is run. In the course of the reaction, heat is either given off or absorbed by the system. Furthermore, the system either does work on its surroundings or has work done on it by its surroundings. Either of these interactions can affect the internal energy of the system. Heat is found on the most basic step by examining the above formula given a specific amount of work and Esystem. The sign of heat corresponds to that of the equation. One can also find ways to make a problem using the equation above more complicated when we tackle problems with constant volume and pressure where we might have to use enthalpy (H).

The amount of heat generated by a reaction can be quantified. The quantity that specifies the amount of heat is known as the “enthalpy” . If is positive then the reaction is endothermic and if it is negative the reaction is exothermic. For example, consider the combustion of methane:

This equation tells us that the combustion of methane is exothermic and releases 883 kilojoules of energy (in the form of heat) for every mole of methane (CH4). Given this information we can determine how much heat will evolve if we burn a certain mass ofmethane: For example: Suppose we combust 20 g of methane, How much heat will evolve (in kJ)? The procedure is the usual stoichiometry:

So we found what heat is and how to find heat quantitatively. Enjoy!

Nnana Amakiri

# Oxidation Number Calculation

One of the most necessary procedures in Chemistry is learning how to calculate oxidation numbers. Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they actually are in any quick way. How do we calculate oxidation numbers?

First off, an oxidation number is the the degree of oxidation of an atom, ion, or molecule; for simple atoms or ions the oxidation number is equal to the ionic charge.

For example, the oxidation number of hydrogen is +1 and of oxygen is -2.

It helps to use a periodic table to determine oxidation numbers.

In between +2 and +3 we do not assign any numbers because there tends to be more than one oxidation number assigned to those elements.

Although the above method is dependable, oxidation states change. Elements can be oxidized or reduced.

Oxidation involves an increase in oxidation state or the decrease in number of electrons

Reduction involves a decrease in oxidation state or the increase in number of electrons

Recognising this simple pattern is the single most important thing about the concept of oxidation states. A few rules to follow when looking for the oxidation number include:

• The oxidation state of an uncombined element is zero. That’s obviously so, because it hasn’t been either oxidised or reduced yet! This applies whatever the structure of the element – whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.
• Some elements almost always have the same oxidation states in their compounds but some, like Hydrogen although usually +1, can be different.

A few examples that you can work out are below.

What is the oxidation state of chromium in Cr2+?

For a simple ion like this, the oxidation state is the charge on the ion – in other words: +2 (Don’t forget the + sign.)

What is the oxidation state of chromium in CrCl3?

This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n:

n + 3(-1) = 0

n = +3 (Again, don’t forget the + sign!)

I hope this helps you as much as it has helped me.

-Nnana Amakiri

# Energy and Temperature Dependence

How does energy and temperature affect reaction rates? I grew interest from this topic when I sought to gain a deeper and more clear understanding of how energy and temperature affect the rate of a reaction.

Consider the reaction from a chemical reference online:

H2 + Cl2 -> 2HCl

On a molecular level, bonds must be broken (H-H and Cl-Cl) before the reaction can proceed too far into products. This means that as the reactant molecules come together, the collision must have enough energy to initiate the bond breakage for the reaction to occur. Not all collisions will have this amount of energy. The collisions that do not have sufficient energy to react end up as elastic scattering events. Dictionary.com shows that Elastic suggests that it is able to maintain its shape or speed after a collision.

Only collisions with enough energy react to form products. The energy of the system changes as the reactants approach each other. The critical amount of energy to make the reaction proceed is called the Activation Energy.

The picture below comes from our chemistry textbook and details the amount of energy needed in a reaction.

From Wikipedia we are told that each reaction rate has a temperature dependency which is usually given by the Arrhenius equation:

$k = A e^{ - \frac{E_a}{RT} }$

Ea is depicted as our activation energy or the amount of energy needed for the reaction to occur. represents the gas constant. represents temperature.

What is important to take from this is that when you conduct a reaction at a higher temperature, you deliver more energy into the system and increase the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy (Ea) resulting in more successful collisions (when bonds are formed between reactants).

Ex:

Coal burns in a fireplace in the presence of oxygen, but it does not when it is stored at room temperature. The reaction is spontaneous at low and high temperatures but at room temperature its rate is so slow that it is negligible. The increase in temperature, as created by a match, allows the reaction to start and then it heats itself, because it is exothermic. That is valid for many other fuels, such as methanebutane, and hydrogen.

Energy and Temperature can greatly influence the presence or the rate of a reaction

Nnana Amakiri