Ch. 10 Kinetics

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Reaction Rates

Example

In the reaction equation CX6HX12OX6+6OX26COX2+6HX2O\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}, suppose the rate of disappearance of CX6HX12OX6\ce{C6H12O6} is xmolCX6HX12OX6Lsx \pu{ mol \ce{C6H12O6}//L s}
Then by mole ratio, we can calculate the rate of appearance of HX2O\ce{H2O} to be
xmolCX6HX12OX6Ls6 molHX2OmolCX6HX12OX6=6xmolCX6HX12OX6Lsx\pu{mol\ce{C6H12O6}//Ls}\cdot\pu{{6 mol \ce{H2O}}//mol\ce{C6H12O6}}=6x\pu{mol\ce{C6H12O6}//Ls}


Factors that Affect Reaction Rates

Example

Given the table of data below, determine the rate law

Experiment
Number
[A][A]
(mol/L)(\pu{mol/L})
[B][B]
(mol/L)(\pu{mol/L})
Initial Rate of Reaction
(mol/Ls)(\ce{mol/Ls})
10.1000.1000.2000.2001.11061.1\cdot10^{-6}
20.1000.1000.6000.6009.91069.9\cdot10^{-6}
30.4000.4000.6000.6009.91069.9\cdot10^{-6}

The general form is Rate=k[A]x[B]y\text{Rate}=k[A]^x[B]^y
Divide the first two equations:
Rate2Rate1=k(0.100)x(0.600)yk(0.100)x(0.200)y\frac{\text{Rate}_2}{\text{Rate}_1}=\frac{k(0.100)^x(0.600)^y}{k(0.100)^x(0.200)^y}
This simplifies to
9.91061.1106=(0.600)y(0.200)y9=3.00yy=2\frac{9.9\cdot10^{-6}}{1.1\cdot10^{-6}}=\frac{(0.600)^y}{(0.200)^y}\rightarrow 9=3.00^y\rightarrow y=2
Next, divide the second two equations:
Rate3Rate2=k(0.400)x(0.600)yk(0.100)x(0.600)y\frac{\text{Rate}_3}{\text{Rate}_2}=\frac{k(0.400)^x(0.600)^y}{k(0.100)^x(0.600)^y}
This simplifies to
9.91069.9106=(0.400)x(0.100)x1=4.00yx=0\frac{9.9\cdot10^{-6}}{9.9\cdot10^{-6}}=\frac{(0.400)^x}{(0.100)^x}\rightarrow 1=4.00^y\rightarrow x=0
Finally, we use any one of the equations to find kk. We use the first experiment:
1.1106=k(0.100)0(0.200)2k=2.81051.1\cdot10^{-6}=k(0.100)^0(0.200)^2\rightarrow k=2.8\cdot10^{-5}
So the final rate law is
Rate=(2.8105 Lmols)[B]2\text{Rate}=(\pu{2.8*10^-5 L//{mol s}})[B]^2


Reaction Order

Zero-Order Reactions

Rate=k\text{Rate}=k
[A]=kt+[A]0[A]=-kt+[A]_0

First-Order Reactions

Rate=k[A]\text{Rate}=k[A]
ln[A]=kt+ln[A]0\ln{[A]}=-kt+\ln{[A]_0}

Second-Order Reactions

Rate=k[A]2 or Rate=k[A][B]\text{Rate}=k[A]^2\text{ or }\text{Rate}=k[A][B]
1[A]=kt+1[A]0, when [A]=[B]\frac{1}{[A]}=kt+\frac{1}{[A]_0}\textit{, when }[A]=[B]


Kinetic Theories


Reaction Mechanisms

Example

What is the reaction rates of the following elementary steps for the reaction 2NO+OX22NOX2\ce{2NO + O2 -> 2NO2}?
NO+OX2NOX3\ce{NO + O2 -> NO3}
NOX3+NO2NOX2\ce{NO3 + NO -> 2NO2}


The first reaction is simply Rate=k[NO][OX2]\text{Rate}=k[\ce{NO}][\ce{O2}]
The second is Rate=k[NOX3][OX2]\text{Rate}=k[\ce{NO3}][\ce{O2}], but since NOX3\ce{NO3} is not in the overall reaction, we must substitute.
From the first equation, kf[NO][OX2]=kr[NOX3][NOX3]=kfkr[NO][OX2]k_f[\ce{NO}][\ce{O2}]=k_r[\ce{NO3}]\rightarrow[\ce{NO3}]=\frac{k_f}{k_r}[\ce{NO}][\ce{O2}]
So, the second is Rate=k(kfkr)[NO][OX2][NO]=k[NO]2[OX2]\text{Rate}=k(\frac{k_f}{k_r})[\ce{NO}][\ce{O2}][\ce{NO}]=k'[\ce{NO}]^2[\ce{O2}]