CHEM 122H Lecture Notes - Lecture 4: Chemical Equation, Joule

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2 Apr 2016

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We shall start of this secion with this very useful equaion: h= u+ ngasrt. A thermochemical equaion is a statement of a chemical equaion and the corresponding reacion enthalpy, the enthalpy change for the stoichiometric amounts of substances in the chemical equaion. For reacions in which no gas is generated or consumed, litle expansion work is done as the reacion proceeds and so the diference between h and u is negligible, so we can set u= h. Using the gas equaion pv=nrt, and taking the diference between enthalpy inal and iniial, we get the equaion h= hinal - hiniial = u + (ninal-niniial)rt. This is how we got the equaion from the irst sentence of these notes. Standard reacion enthalpies: because the heat released or absorbed by a reacion depends on the physical states of the reactants and products, when calculaing the reacion enthalpy we need to specify the state of each substance.

Related Questions

Â± Relating Different Forms of the Equilibrium Constant

For chemical reactions where all reactants and products are in the gas phase the amount of each gas in the vessel can be expressed either as partial pressures or as concentrations. As such the equilibrium constant for a gas phase reaction can also be expressed in terms of concentrations or pressures. For the general reaction,

aA(g)+bB(g)âcC(g)+dD(g)

Kp=(PC)c(PD)d(PA)a(PB)b and Kc=[C]c[D]d[A]a[B]b.

It is possible to interconvert between Kp and Kc using

Kp=Kc(RT)În

where R=0.08314 L bar molâ1 Kâ1 and În is the difference in stoichiometric coefficients between gaseous products and gaseous reactants.

If we assume that the gaseous reactant and products are âidealâ and the partial pressures of the reactants and products were expressed in bar, then the thermodynamic equilibrium constant, K is the magnitude (or numerical value) of Kp.

Part A

For the reaction

2CH4(g)âC2H2(g)+3H2(g)

Kc=0.160 mol2 Lâ2 at 1621 âC. What is Kp for the reaction at this temperature?

Enter your answer numerically.

Part B

For the reaction

N2(g)+3H2(g)â2NH3(g)

Kp=3.80Ã10â3 barâ2 at 265 âC. What is Kc for the reaction at this temperature?

Enter your answer numerically.

M 121 1 201 Ideal Gas Law (PV n Experiment to T ) Determine the Ideal Gas Constant R 1. Conduct basic measurements, measurement , mathematical calculations and conversions for of volume, pressure, and temperature: calculate percent error and account for sources of error in an e xperimental result name and by following safety proced ty experiments for a redox reaction by using equipment identifiable by form chemical calculations using mass, mole and volumes and idenify limitatios of measuring devices in order to state the uncertainty/significant figures in measurements and calculation results. 5. Demonstrate a proficiency in relating the chemistry of gases to everyday phenomenon. Introduction Molecules in gases have very little attraction for one another, because relative to their size, they are so far apart from each other. For this reason we can often assume that a gas behaves "idcally." meaning that the physical behavior of the gas is independent of the composition of the gas. If a gas behaves ideally, then its physical properties can be described by the Ideal Gas Equation: Let's consider why the variables in the ideal gas equation have the relationships expressed in Equation 1aã The ideal gas equation assumes that there are no attractive forces between gas molecules. One can picture two billiard balls bouncing off one another in a totally clastic collision. If these molecules are placed in a balloon the collisions will occur not only with each other but also with the wall of the balloon, causing the balloon to expand. This expansion can be measured as a volume (V). More molecules will cause more collisions and thus more expansion. Therefore the quantity of molecules or number of moles (n) is directly proportional to the volume, and thus volume and number of moles appear on opposite sides of the equation. Another factor to consider is the kinetic energy of these molecules. Kinetic energy is related to the temperature (T) of the molecules, in that the kinetic energy. If the collisions with the wall of the balloon are happening at a very high speed due to high kinetic energy, the collisions will be harder and more frequent and thus the wall of the balloon will be pushed out further (a larger volume) than if the kinetic energy were less. The result is that volume is also directly proportional to the temperature. This relationship is seen in Equation la with (T) on the right side of the equation and (V) on the left. higher the temperature the greater the considered. While the molecules inside the balloon are pushing the balloon out, there are also molecules in thea Finally the pressure of the outside environment (atmospheric pressure) needs to be 59

CHEM 122H Lecture Notes - Lecture 4: Chemical Equation, Joule

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