Experiment No. | [S2O8] (mol/L) | [I-] (mol/L) | Initial Rate of Rxn |
---|---|---|---|
1 | .2 | .2 | 2.2x10^-3 |
2 | .4 | .2 | 4.4x10^-3 |
3 | .4 | .4 | 8.8x10^-3 |
This makes the rate law Rate = k[S2O8]^x[I-]^y
Experiment 1 and 2 can be used to find the exponent for S2O8 because I- is held constant
Experiment 3 and 2 can be used for [I-] exponent
To then find k, use the values from any ONE experiment and solve. any experiment values can be used.
Order of reaction is used to indicate the exponents in the rate law
The half life of Pb 210 is 25 years. In a sample of 50 µg of Pb 210, how much is left after 100 years
How many years of 50 µg of Pb 210 takes to decrease to 5.0µg ( t(½) =25yrs)
Rate = k[A]^2
Rate = k [A][B]
Integrated rate law
Half-life equation
According to the transition state theory, as molecules approach each other, they slow down and potential energy increases. After they get as close as they can, the potential energy decreases and kinetic increases.
The minimum amount of kinetic energy that must be converted is the activation energy
Reaction profiles help determine if a reaction is endothermic or exothermic.
Heat of reaction (ΔH) can be determine from potential energy of product and reactants
If ΔH is positive, heat is absorbed and is endothermic
If ΔH is negative, heat is release and is exothermic
Catalysts allows a reaction to speed up by providing an alternate reaction pathway with a lower activation energy. However, this does not change the amount of product formed or the equilibrium concentrations.
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