Energy Calculations ( The return of the MOLE(>o<))

ΔH is the energy change that takes place in an reaction and is expressed in kJ/mole. It can be on both sides of a chemical equation. If it's on the left hand side (LHS) of the equation which is terminologically called reactants, it's a endothermic reaction that loses energy. If it's on the right hand side (RHS), the reaction, it's called exothermic.

Now, in order to do a energy calculation, one just need to use the ΔH/the coefficients of the balanced equations to find out how much energy released/absorbed from the substances.

EX. C3H8 + 5O2 -----> 3CO2 + 4H2O + 1050kJ
       so,   __-1050kJ___          or          __-1050kJ__  = __-210kJ__
              1 mol of C3H8                         5 mols of O2      1 mol of O2

NOTE: ΔH (energy change) differentiates base on what chemical reaction one is preceeding.

Examples of more advance energy calculations:

Endothermic V.S Exothermic Reactions

Today in Class we lernt about Exo and Endo Reactions.  We also lernt about  their diagrams.  The Important concept behind the two distinct diagrams is the Reactant, Product, Activation Energy, Change in Enthalpy, Activated Complex and the Change in Enthalpy.  Therefore these are the things we need to focus on in this lesson.




Endothermic reactions: the products have more energy than the reactants. Heat is absorbed from the surroundings (e.g., a cold pack).

Reactant:  The reacant will have lower energy than the product since energy is absored during the reaction

Product:  The product will gain energy.  Thus have more energy than before

Change in Enthalpy: Delta H>0

Exothermic reactions: the products have less energy than the reactants. Heat is released to the surroundings (e.g., a heat pack).

Reactant:  The reactant will have higher energy than before since energy is released during the reaction.

Product:  The product will  have lower energy than before because heat is released.

Change in Enthalpy: Delta H<0

Activation Energy:  It is energy required from the reactant to make it to the tob of the Activation complex.  This can be calculated by energy of the product - the energy of the reactant.

Activation Complex:  It is the peak of the porabola on the energy diagram.  It is located usually in the middle of the graph

Change in Enthalpy:  It is change in energy between the reactant and the product.  This is calcuated by useing the energy of the product to minus the energy of the reactant.



Both reactions require some activation energy to begin the reaction.

Lab 5B

 Today in Chemistry, we lernt to classify the four common chemical reaction through experiment. Lab 5 B helps gives us a better understanding of what each reactions will look like.

Reactions

Reactions#1:  We use a crucible tongs to hold a 6cm copper wire on the hottest part of the flame for serveral minutes.  In the end we define it as Synthesis because Oxgen reacted during this process with copper.  Thus it together formed Copper Oxide.

Reaction#2:  We place an iron nail in a test tube filled with Copper two sulfate solution.  We waited for 15min and in the end they both reacted and changed color.  To conclude we define the reaction as single replace ment because the iron during was able to replace Copper and formed Iron Sulfate.

Reaction#3:  We put Solid Copper two Sulfate Pentahydrate in the testtube.  Than we move the test tube back and forth gently over a burner flame.  In the end we define that reaction as deomposion because the heat absorded during the reaction was able to break the bond and formed seperate Copper two sulfate and Pentahydrate.

Reaction#4:  We extend the process of reaction 3 by putting a few drops of water back into the test tube.  During this process the Solid Copper two sulfate was able to gained back its Pentahydrate.  Thus Synthesis occured.

Reaction#5:  We filled test Tube with Calcium Chloride Solution and we mixed it with Sodium Carbonate Solution.  At the end we define the reaction as Double Replacement because Calcium exchanged with Sodium to form Calcium Carbonate and this goes same for Sodium who exchanged to form Sodium Chloride.

Reaction#6:  We place a pice of mossy zinc in test tube to react with Hydrochloric acid solution.  In the end we define the reaction as Single replacement because the zinc replace hydrogen and formed zinc chloride leaving hydrogen gas alone.

Reaction#7:  We filled test tube with hydrogen peroxide solution than we add manganese four oxide. In the end we define the reaction as decompositon because both hydrogen and oxgen serparated from hydrogen peroxide.  We test the gas by placing a glowing splint into the testtube moth .

That Concludes Lab 5B.

Virtual Lab

       Today in Chemistry we did a virtual lab in the computer lab.  The purpose of the lab is to let us  review the concept of the activity series use for identifying single replacement reactions.  We use different examples of metals to form different kinds of ions throughout the lab.  The pattern we discovered at the end is that the higher reactive metals forms lower reactive ions.  It is because the new formed ion by the highly reactive metal is harder to be replace by another metal during single replacement.  Thus there is lower chance for it to  become reactive again unless replace by a higher reactive metal than before.

The metals we use during reaction: Ag, Zn, Mg, Fe, Pb, Ni  and Sn