2.  The chemical equation does no tell us anything about the conditions required for the reaction to occur. Also it does not give any information about the speed or rate at which the reaction takes place.  Rate of reaction means the speed at which the reaction takes place.

3. 1. Chemical nature of reactants (some chemicals have a tendency to react more than other chemicals) 2. Ability of reactants to make contact (the phase of reactants. Chemicals can react quickly when both are gases, liquids of aqueous because molecules/ions can intermingle. Solids reacts slowly because molecules don not mix) 3. Temperature

4. 4. Concentration (a high concentration  more molecules to react  greater no. of collision/second  increasing the rate) 5. Presence of catalyst (a substance that speeds a chemical reaction, but is not itself alerted or used up during the reaction)

5.  Homogeneous reaction: the reactants are in the same phase  Heterogeneous reaction: the reactants are in separate phases.

6.  At a fixed temperature, the rate of a reaction may be mathematically described by a rate law. Consider the following hypothetical reaction: A + 2B + 3C  product  The differential rate law for this reaction represent the rate of change in concentration of a reactant and has the following form Rate = α [A] x [B] y [C] z Rate = K [A] x [B] y [C] z (mol L -1 s -1 ) where K is the rate constant, x, y, and z are the order with respect to each reactant. The over all order of the reaction is the sum of x, y, and z.

7.  In this experiment hydrogen peroxide slowly oxidizes iodide ion to iodine 2I - + 2H + + H 2 O 2  I 2 + 2H 2 O Rate = K [I - ] x [H 2 O 2 ] y [H + ] z  the concentration of H+ is held constant throughout the experiment then its effect will not appear in the rate law. R = K [I - ] x [H 2 O 2 ] y

8.  In this experiments sodium thiosulfate is used as a catalyst.  Starch is added in this experiment to indicate the presence of iodine. Used as indicator. I 2 + starch  blue complex

9. 0.1M H 2 O 2 Solution A Test no. 0.02M Na 2 S 2 O 3 Starch soln. 0.3M KIBuffer soln. Distilled water 302521025701 30202 25752 301523025803 501023025854 70523025905

10. K=R/[I - ] x Log RR=[I - ]/tLog [I - ]M KIM H 2 O 2 Test You have to find it 0.11 You have to find it 0.12 You have to find it 0.13 Part A: at constant concentration of H 2 O 2: To find the concentration of KI: From chemical equation: 2 I - + H 2 O 2  I 2 + 2H 2 O M I - = 2MV (H 2 O 2 ) /V (KI)

11. Part B: at constant concentration of I - K=R/[H 2 O 2 ] y Log R R=[H 2 O 2 ]/t Log [H 2 O 2 ] M KIM H 2 O 2 Test 0.3You have to find it 3 0.3You have to find it 4 0.3You have to find it 5 From the chemical equation find the relation between the 2 reactants and find the concentration of H 2 O 2

12.  From part A plot a graph between Log R(y axis) and Log [I - ] (X axis) and find the slope  The slope will equal the value of x.  Substitute x in the table to find K.  Do the same with part B to find y.  At the end get the average K for the reaction.