In: Chemistry
Pre-Lab Questions
experiment 1: temperature and the rate of chemical reaction
Data Sheet
Table 1. Temperature vs. Brightness of Glow Stick
Beaker |
Temperature (ºC) |
Brightness Before Activation |
Brightness After Activation |
Start Time |
Stop Time |
Beaker 1 (Room Temp) |
|||||
Beaker 2 (Cold Water) |
|||||
Beaker 3 (Hot Water) |
Graph of Your Data:
Post-Lab Questions
* 1. Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.
2.Temperature. An increase in temperature typically increases the rate of reaction. An increase in temperature will raise the average kinetic energy of the reactant molecules. Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision .
* The glow stick contains two chemicals, a base catalyst, and a suitable dye (sensitizer, or fluorophor). This creates an exergonic reaction. The chemicals inside the plastic tube are a mixture of the dye, the base catalyst, and diphenyl oxalate. The chemical in the glass vial is hydrogen peroxide.
*A chemical reaction is usually accompanied by easily observed physical effects, such as the emission of heat and light, the formation of a precipitate, the evolution of gas, or a color change. Absolute confirmation of a chemical change can only be validated by chemical analysis of the products.
*This allows the inside the glass to mix with the chemicals outside the glass vial. Once these chemicals contact one another, a reaction takes place. The chemical reaction in a glow stick releases energy in the form of light. This light along with dye allows the stick to glow the brilliant colors that we observe.
* Decomposition of H2O2 is a first order reaction. So the rate = K[H2O2]
Now we assume initially the concentration of H2O2 is 'X' M.
So rate =K × X
Now when the concentration is double it's original value i. e. [H2O2] = 2X Then,
Rate = K[H2O2] =K ×2X=2KX
Therefore if the concentration of H2O2 is double it's original value the rate of the reaction will be double.