Stoichiometry,+Gas+Laws

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 * Gas Laws and Stoichiometry

Ideal Gas Law ** **Partial Pressures** **Kinetic Theory**

Gases expand unifromly to fill any container in which it is placed. The volume of the gas is the volume of the container. Gases also have pressure. Pressure is defined as force per unit area. Barometers are used to measure pressure and is often expressed in millimeters of mercury. This can also be converted to atmospheres (atm). Use the following conversion factors:
 * WHAT IS A GAS?**

(10^-3m^3)/1L (4.003g He)/1mol He (760mmHg)/1atm (1.013bar)/1atm

**__The Early Gas Laws __**
 CONTRIBUTION MADE BY CAMILLE SHARPE

The early gas laws were developed at the end of the eighteenth century, when scientists began to realize that relationships between the pressure, volume and temperature of a sample of gas could be obtained which would hold for all gases. Gases behave in a similar way over a wide variety of conditions because to a good approximation they all have molecules which are widely spaced, and nowadays the equation of state for an ideal gas is derived from kinetic theory. The earlier gas laws are now considered as special cases of the ideal gas equation, with one or more of the variables held constant. Three earlier gas laws: , -graph~Katelyn > -graph~Katelyn
 * Boyle's law (1662, relating pressure and volume):
 * Charles' law or law of volumes (1787, relating volume and temperature):


 * Pressure law or Third gas law (Gay-Lussac in 1809, relating temperature and pressure)

-graph~Katelyn

The combined gas law or general gas equation is formed by the combination of the three laws, and shows the relationship between the pressure, volume and temperature for a fixed mass of gas: http://en.wikipedia.org/wiki/Gas_laws

**THE IDEAL GAS LAW ** All gases closely resemble each other in the dependence of volume on amount, temperature, and pressure 1. Volume is directly proportional to amount 2. Volume is directly proportional to temperature (in kelvin) 3. Volume is inversely proportional to pressure.

PV=nRT P=pressure V=volume n=moles R= .0821 L*atm/(mol*K)
 * There are a few variations of PV=nRT: MM= gRT/PV, D= (MM)(P)/RT, P1V1/n1T1 = P2V2/n2T2 (contribution by Mr. Daddio)

The ideal gas law can be used to: 1. Find the final state of a gas, knowing its inital state and the changes in its P,V, N, or T 2. One of the four missing variables (such as what is the temperature of the gas in kelvin given the other information about the gas?) 3. The molar mass or density of a gas

[] This link demonstrates how changing each variable effects the gas molecules.

There will be an Example problem using PV=nRT!!! Santo-

**REFER TO POWERPOINT FOR AN EXAMPLE PROBLEM INVOLVING THE IDEAL GAS LAW**

Stoichiometry in gas The first relationship discovered was the law of combining volumes, made by Gay-Lussac. He said that the volume ratio of any two gases in a reaction at constant temperature and pressure is the same as the reacting mole ratio.

For example: 2H2(g) + O2(g)>2H20(l) a) What volume of H2(g) at 25 degrees celsius and 1.00 atm is required to react with 1.00L of O2(g) at the same temperature and pressure? b) What volume of H20(l) at 25 degrees celsius and 1.00atm (d=.997g/mL) is formed from the reaction in (a)? c) What mass of H20(l) is formed from the reaction in (a), assuming a yeild of 85.2%?

a) according to the law of combinging volumes, volumeH2 = 1.00L O2 x (2 volumes H2/ 1 volume O2) = 2.00L H2 b) nO2g = PV/RT= ((1.00atm)(1.00L))/((0.0821L *atm/mol*K)(298K) = 0.0409 mol O2 nH20= 0.0409 mol O2 x (2mol H20/1 mol O2)= 0.0818 mol H20 VH20= 0.0818 mol H20 x (18.02 g H20/ 1 mol H20) x (1.00mL/ 0.997g) = 1.48mL c) The theoretical yield of H20 as calculated in (b) is 0.0818 mol. It follows that yield H20 = 0.0818mol H20 x (18.02g H20/1mol H20) = 0.852 = 1.26 g H20

For a mixture of two gases A and B, the total pressure is given by the expression Ptot = ntot(RT/V) = (nA + nB)(RT/V) Ptot = PA + PB Dalton's Law: The total pressure of a gas mixture is the sum of the partial pressures of the components of the mixture.
 * GAS MIXTURES: PARTIAL PRESSURES AND MOLE FRACTIONS **

The partial pressure of water vapor, PH20, is equal to the vapor pressure of the liquid pressure of liquid water. The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. PA = XaPtot


 * KINETIC THEORY OF GASES **
 * Gases are mostly empty space
 * Gas molecules are in constant, chaotic motion.
 * Collisions are elastic.
 * Gas pressure is caused by collisions of molecules with the walls of the container.

Molecules will flow out of these leaks, in a process called [|effusion]. Because massive molecules travel slower than lighter molecules, the rate of effusion is specific to each particular gas. We use **Graham's law** to represent the relationship between rates of effusion for two different molecules. This relationship is equal to the square-root of the inverse of the molecular masses of the two substances.



Where: r1=rate of effusion in molecules per unit time of gas "1" r2=rate of effusion in molecules per unit time of gas "2" u1=molecular mass of gas "1" u2=molecular mass of gas "2"

1) Write a balanced equation for the reaction. Once all quantities have been converted to moles and you have a balanced reaction, you are ready to actually use stoichiometry**. The idea of stoichiometry is really quite simple. The coefficients in front of each reactant and product in the balanced chemical reaction tells you the ratio of how much of each you will react/produce. Let's take a simple example: the reaction of hydrogen gas (H2) with oxygen gas (O2) to form water (H2O). 2H2 + O2 ---> 2H2O In the balanced reaction, there is a 2 in front of the H2, and although nothing is written, that means that there is really 1 in front of O2, and a 2 in front of H2O. **Stoichiometry tells us that because of the way the numbers in balanced reaction came out, we need two molecules of H**2 **to react with each one molecule of O**2**, and also, that this will form 2 molecules of water.** I can say the same thing using moles: For each mole of O2 I react, I need 2 moles of H2, and I will produce 2 moles of O2. All I'm using is the ratio of the coefficients from the balanced reaction. That is stoichiometry! So, using this I can say things like: -- If I reaction 0.5 moles of O2 completely, I will make 1 mole of water -- If I made 4 moles of water, then I consumed 4 moles of H2 and 2 moles of O2. -- If I want to completely react 5 moles of H2, then I need 2.5 moles of O2 (and I will get 5 moles of H2O). Makes sense? The ratio of H2 to O2 to H2O is 2:1:2, and that always holds for this reaction. --Problem: If I burn 10 grams of methane, how many grams of CO2 will be produced? -- Answer: As I said, to solve this problem we need two things: a balanced reaction, and to convert all quantities to moles. First, let's write a balanced reaction. In this reaction, methane gas (CH4) gets burned in oxygen (O2) to form carbon dioxide (CO2) and water vapor (H2O). The balanced reaction is: CH4 + 2O2 --> CO2 + 2H2O Notice the ratio is now 1:2:1:2 (reading the reaction from left to right). So that tells me for every mole of CH4 I react, I need 2 moles of O2, and I will get out 1 mole of CO2 and 2 moles of H2O. Now, I need to convert the 10 grams of methane into moles, because **stoichiometry only works for moles and NOT grams**! So I use the molar mass of CH4, which is 12.011+4*1.0079 = 16.0426 grams per mole. So to convert to moles, I just divide: 10 grams ÷ 16.0426 gram/mole = 0.6233 moles From the stoichiometry, I now know that if I react 0.6233 moles of methane, I will need twice that many moles of oxygen, or 1.2467 moles O2, and I will get 0.6233 moles of CO2 and 1.2467 moles of H2O as products. But the question asked for a number of grams of CO2, not moles. So once I'm done using stoichiometry, I convert back to grams, now using the molar mass of CO2 (which is 12.011 + 2*15.999 = 44.009 g/mol). So 0.6233 moles of O2 is: 0.6233 moles * 44.009 grams/mole = 27.4308 grams of CO2 Notice that the ratio of moles of CH4 to CO2 is 1 to 1, but the ratio of the weights is totally different. Remember, **STOICHIOMETRY ONLY WORKS ON MOLES**! I can also find how many grams of water I'll produce, just for fun! The molar mass of water is 15.999 + 2*1.0079 = 18.0148 g/mol. So to convert 1.2467 moles H2O to grams: 1.2467 moles * 18.0148 gram/mole = 22.4591 grams of H2O []
 * Stoichiometry Problems **
 * 2) Convert all amounts of products and/or reactants in the question into moles.
 * Now let's make it a bit more complicated.**



Example problem using Ideal Gas Law "PV=nRT" [|chemstuff.pdf]

Example problem using stoichiometry and Ideal Gas Law "PV=nRT" [|chemstuff2.pdf]