Chemical Kinetics The Second Order Integrated Rate Equation

Second Order Rate Law CHEM 3310

This is the second order differential rate law. Second Order (m+n=2) For a general reaction that obeys second order rate law. a X + b Y  products There are a number of ways for m+n=2. The most simple is This is the second order differential rate law. The rate of a second order reaction is proportional to the concentration of the reactant squared. Increasing the concentration of the reacting species will increase the rate of the reaction. (i.e. Doubling concentration increases the rate 4x.) CHEM 3310

This is the second order differential rate law. Second Order (m+n=2) For a general reaction that obeys second order rate law. a X + b Y  products There are a number of ways for m+n=2. The most common is This is the second order differential rate law. When this equation is rearranged and both sides are integrated we get the following result. CHEM 3310

that obeys second order rate law. For a general reaction that obeys second order rate law. a X + b Y  products This is the second order differential rate law. Separate the variables Integrate This is the second order integrated rate law.

This is the second order integrated rate law. For a general reaction that obeys second order rate law. a X + b Y  products This is the second order integrated rate law. [X]0 is the concentration of X at time=0 [X]t is the concentration of X at time=t A reaction is second order if the concentration of X decreases inversely with time. A plot of is a straight line. The slope of this resulting line is ak2 The unit of k2 is M-1s-1 CHEM 3310

that obeys second order rate law. For a general reaction that obeys second order rate law. a X + b Y  products Integrated rate law Differential rate law CHEM 3310

that obeys second order rate law. For a general reaction that obeys second order rate law. a X + b Y  products Integrated rate law Differential rate law A graph of “[X]-1 versus time” would yield a straight line. Slope = ak2 CHEM 3310

The more concentrated the reactant is, the shorter the half-life. Second Order For a general reaction that obeys second order rate law. a X + b Y  products Integrated rate law Half-life,  When t= , Half-life of a second-order reaction depends on the inverse of the initial concentration of the reactant. The more concentrated the reactant is, the shorter the half-life. To test a set of data, could plot “ versus [X]o-1” to see if it yields a straight line with slope = (ak2)-1.

that obeys second order rate law. Second Order (m+n=2) For a general reaction that obeys second order rate law. a X + b Y  products Overall second order; First order with respect to X; First order with respect to Y. The integrated form is complicated. In Experiment 3, CH3COOC2H5 (aq) + OH- (aq)  CH3COO- (aq) + C2H5OH (aq) Since the reaction is 1:1, we adjust the initial concentration of to be the same as that of CH3COOC2H5. Rate = k [CH3COOC2H5] [OH-] Then, the rate equation reduces to Rate = k [CH3COOC2H5]2 or Rate = k [OH-] 2 CHEM 3310

Show that this reaction is 2nd order and not 1st order. Second Order Example: The decomposition of nitrogen dioxide 2 NO2 (g)  2 NO (g) + O2 (g) Data collected in this kinetic study: [NO2] (M) Time (s) 1.00 x 10-2 8.00 x 10-3 50 6.47 x 10-3 100 4.76 x 10-3 200 3.74 x 10-3 300 Show that this reaction is 2nd order and not 1st order. CHEM 3310

Second Order Example: The decomposition of nitrogen dioxide 2 NO2 (g)  2 NO (g) + O2 (g) Raw data: What would you plot to test what order this reaction is? Is it first-order? Is it second-order? [NO2] (M) Time (s) 1.00 x 10-2 8.00 x 10-3 50 6.47 x 10-3 100 4.76 x 10-3 200 3.74 x 10-3 300 Graph “ln [NO2] versus time” to see if it yields a straight line. Graph “1 / [NO2] versus time” to see if it yields a straight line. CHEM 3310

Second Order Example: The decomposition of nitrogen dioxide 2 NO2 (g)  2 NO (g) + O2 (g) Original data ln [NO2] Time (s) -4.605 -4.828 50 -5.041 100 -5.348 200 -5.589 300 To test for 1st order: Take the natural log of [NO2] and plot it against time. [NO2] (M) Time (s) 1.00 x 10-2 8.00 x 10-3 50 6.47 x 10-3 100 4.76 x 10-3 200 3.74 x 10-3 300 To test for 2nd order: Take the inverse of [NO2] and plot it against time. 1/[NO2] (M-1) Time (s) 100. 125. 50 155. 100 210. 200 267. 300 CHEM 3310

Second Order Example: The decomposition of nitrogen dioxide 2 NO2 (g)  2 NO (g) + O2 (g) The “ln [NO2] versus Time” graph for this reaction does not yield a straight line, proving that it is not a first order reaction. The “1 / [NO2] versus Time” graph for this reaction yields a straight line, proving that it is a second order reaction. CHEM 3310

Concentration of HI (M) Example: We study the decomposition of hydrogen iodide at 508oC, by Measuring the rate at which either H2 or I2 is formed, or the rate at which HI is consumed. 2 HI(g)  H2(g) + I2(g) Experimental Data: Time (s) Concentration of HI (M) 0.100 50 0.0716 100 0.0558 150 0.0457 200 0.0387 250 0.0336 300 0.0296 350 0.0265 Test the data to determine the order of the reaction. CHEM 3310

2 HI(g)  H2(g) + I2(g) Plot “[HI] vs. Time” to test for zeroth order R² = 0.8649 Plot “[HI] vs. Time” does not yield a straight line. The decomposition of HI is not a zeroth order reaction. CHEM 3310

2 HI(g)  H2(g) + I2(g) Plot “ln [HI] vs. Time” to test for first order R² = 0.965 Plot “ln [HI] vs. Time” does not yield a straight line. The decomposition of HI is not a first order reaction. CHEM 3310

2 HI(g)  H2(g) + I2(g) Plot “1/[HI] vs. Time” to test for second order Plot “1/[HI] vs. Time” yields a straight line. The decomposition of HI is a second order reaction. CHEM 3310

Therefore, rate of disappearance of HI is We study the decomposition of hydrogen iodide at 508oC, by measuring the rate at which either H2 or I2 is formed, or the rate at which HI is consumed. 2 HI(g)  H2(g) + I2(g) Experimentally we find that this is a 2nd order reaction. Question: Calculate the rate at which HI disappears in the following reaction at the moment when I2 is being formed at a rate of 1.8 x 10-6 M sec-1. Answer: Therefore, rate of disappearance of HI is 2(1.8 x 10-6) = 3.6 x 10-6 Msec-1 CHEM 3310

Summary of the Rate Laws CHEM 3310

The Rate Law tells us the instantaneous rate (the slope of the curve) as a function of concentration. CHEM 3310

For a general reaction, a X + b Y  products Zeroth Order First Order Second Order Differential Rate Law Integrated Rate Law Units of k M s-1 s-1 M-1 s-1 Linear Plot [X] vs. t ln([X]) vs. t 1/[X] vs. t Half-life CHEM 3310

The Integrated Rate Law tells us how the concentration behaves as a function of time. CHEM 3310