Rates of Chemical Reactions 1: A Clock Reaction 5. From your knowledge of x and y in the equation (as well as the rate in a given experiment from your graph), calculate k from your data. Rate = k[S2O8-2]x[I-]y 5. From your knowledge of x and y in the equation (as well as the rate in a given experiment from your graph), calculate k from your data. Rate = k[S2O8-2]x[I-]y
The Correct Answer and Explanation is:
To calculate the rate constant kk for the clock reaction, we use the rate law: Rate=k[S2O82−]x[I−]y\text{Rate} = k[\text{S}_2\text{O}_8^{2-}]^x[\text{I}^-]^y
Step-by-Step Solution
Assume from previous experiments or the graph you determined:
- The order with respect to persulfate x=1x = 1
- The order with respect to iodide y=1y = 1
(This is typical for the iodine clock reaction if confirmed by experimental data.)
From one experiment (for example):
- [S2O82−]=0.020 M[\text{S}_2\text{O}_8^{2-}] = 0.020 \text{ M}
- [I−]=0.040 M[\text{I}^-] = 0.040 \text{ M}
- Rate = 4.0×10−5 mol/L\cdotps4.0 \times 10^{-5} \, \text{mol/L·s}
Plug into the Rate Law:
4.0×10−5=k(0.020)1(0.040)14.0 \times 10^{-5} = k(0.020)^1(0.040)^1 4.0×10−5=k(8.0×10−4)4.0 \times 10^{-5} = k(8.0 \times 10^{-4}) k=4.0×10−58.0×10−4=0.050 L/mol\cdotpsk = \frac{4.0 \times 10^{-5}}{8.0 \times 10^{-4}} = 0.050 \, \text{L/mol·s}
Explanation:
In chemical kinetics, the rate law expresses how the rate of a reaction depends on the concentration of its reactants. For the iodine clock reaction, the commonly studied reaction involves persulfate ions (S2O82−)(\text{S}_2\text{O}_8^{2-}) oxidizing iodide ions (I−)(\text{I}^-) to form iodine. The rate law for this reaction is empirically determined and written as: Rate=k[S2O82−]x[I−]y\text{Rate} = k[\text{S}_2\text{O}_8^{2-}]^x[\text{I}^-]^y
The exponents xx and yy represent the reaction order with respect to each reactant and must be determined experimentally. In typical iodine clock reactions, both are found to be first order (x=y=1x = y = 1).
Once the orders are known, you can determine the rate constant kk using experimental data for a specific trial. You need the concentrations of the reactants and the measured rate of the reaction. By substituting the known values into the rate equation and solving for kk, you find the proportionality constant that links concentration and rate for the given conditions.
In our example, with given concentrations and a measured rate, we calculated: k=Rate[S2O82−][I−]=4.0×10−5(0.020)(0.040)=0.050 L/mol\cdotpsk = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{4.0 \times 10^{-5}}{(0.020)(0.040)} = 0.050 \, \text{L/mol·s}
This value of kk is crucial for predicting reaction rates under different conditions and for comparing the speed of this reaction to others
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