Chemical Kinetics The First Order Integrated Rate Equation

First Order Rate Law CHEM 3310

This is the first order differential rate law. First Order (m+n=1) For a general reaction that obeys first order rate law. a X + b Y  products This is the first order differential rate law. The rate of a first order reaction is proportional to the concentration of the reactant. Increasing the concentration of the reacting species will increase the rate of the reaction. (i.e. Doubling concentration double the rate.) CHEM 3310

This is the first order differential rate law. For a general reaction that obeys first order rate law. a X + b Y  products This is the first order differential rate law. When this equation is rearranged and both sides are integrated we get the following result. CHEM 3310

that obeys first order rate law. For a general reaction that obeys first order rate law. a X + b Y  products This is the first order differential rate law. Separate the variables Integrate This is the first order integrated rate law. CHEM 3310

What would the “log [X] versus t” graph look like? First Order For a general reaction that obeys first order rate law. a X + b Y  products Integrated rate law Differential rate law Take natural log of both sides, A graph of “ln [X] versus time” would yield a straight line. What would the “log [X] versus t” graph look like? Slope = -ak1 CHEM 3310

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

This is the first order integrated rate law. For a general reaction that obeys first order rate law. a X + b Y  products This is the first 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 first order if the concentration of X decreases exponentially with time. A plot of or is a straight line. The slope of this resulting line is –ak1 The unit of k1 is s-1 The quantity, is the fraction of X that remains at time t. CHEM 3310

   First Order For a general reaction that obeys first order rate law. a X + b Y  products Integrated rate law Half-life,  When t= ,    CHEM 3310

that obeys first order rate law. For a general reaction that obeys first order rate law. a X + b Y  products Integrated rate law Half-life,  When t= , Half-life of a first-order reaction depends only on the rate constant, k1, and independent of the initial concentration of the reacting species. CHEM 3310

First Order Example: The decomposition of dinitrogen pentoxide N2O5 (g)  2 NO2 (g) + ½ O2 (g) The “Concentration versus Time” graph and the “Rate versus Concentration” graph for this reaction is shown below: CHEM 3310

Rate = k [N2O5] Rate = (1.00 x 10-5sec-1)(0.25 M) The decomposition of dinitrogen pentoxide, N2O5 , at a certain temperature has a rate constant, k=1.00 x 10-5 sec-1. N2O5 (g)  2 NO2 (g) + ½ O2 (g) 1. Write the rate law for this reaction. Rate = k [N2O5] 2. What is the rate if the concentration of N2O5 is 0.25 M? Rate = (1.00 x 10-5sec-1)(0.25 M) = 2.5 x 10-6 moles liter-1sec-1 3. If the initial concentration of N2O5 is 0.50 M, what will its concentration be after 100. hours? [N2O5]t=1.4 x 10-2 M CHEM 3310

is the quantity that remains. The decomposition of dinitrogen pentoxide, N2O5, at a certain temperature has a rate constant, k = 1.00 x 10-5 sec-1. N2O5 (g)  2 NO2 (g) + ½ O2 (g) is the quantity that remains. CHEM 3310

What is the order of this reaction? The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) What is the order of this reaction? Method 1: We could measure the volume of O2 (g) evolved as a function of time and relate this volume to the decrease in concentration of H2O2. O2 H2O2 in a sealed vessel. CHEM 3310

set of data to convince yourself that this reaction proceeds via The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) What is the order of this reaction? Method 2: Remove small samples at various times and analyze for H2O2 by titration with KMnO4. 2 MnO4- (aq) + 5 H2O2 (aq) + 6 H+ (aq)  2 Mn2+ (aq) + 8H2O (l) + 5 O2 (g) Time (s) [H2O2] (M) 2.32 300 1.86 600 1.49 1200 0.98 1800 0.62 3000 0.25 Graph this set of data to convince yourself that this reaction proceeds via 1st order! CHEM 3310

What is the order of this reaction? The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) What is the order of this reaction? Raw data: What would you plot to test what order this reaction is? Is it zeroth-order? Is it first-order? Time (s) [H2O2] (M) 2.32 300 1.86 600 1.49 1200 0.98 1800 0.62 3000 0.25 Graph “[H2O2] versus time” to see if it yields a straight line. Graph “ln [H2O2] versus time” to see if it yields a straight line. CHEM 3310

What is the order of this reaction? The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) What is the order of this reaction? Graph “[H2O2] vs. time” to test for 0th order! Graph “ln [H2O2] vs. time” to test for 1st order! rate constant, k A straight line indicates that the decomposition of H2O2 is 1st order. Rate constant, k, is 7.4x10-4 s-1 CHEM 3310

Make sure you can solve using your calculator. The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) First order reaction; k = 7.4x10-4 s-1 The initial concentration of H2O2 is 2.32 M. (a) What will be the [H2O2] at t = 1200 s? Use the first order integrated equation. [H2O2]t = ? [H2O2]0= 2.32 M t = 1200 s k = 7.4x10-4 s-1 Make sure you can solve using your calculator. [H2O2]t =0.95 M 1200 s CHEM 3310

Make sure you can solve this equation using your calculator. The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) First order reaction; k = 7.4x10-4 s-1 The initial concentration of H2O2 is 2.32 M. (b) What percent H2O2 has decomposed in the first 500 s after the reaction begins? Use the first order integrated equation in logarithmic form. represents the fraction of the H2O2 that remains unreacted at time t. represents the fraction of the H2O2 that has decomposed at time t. Make sure you can solve this equation using your calculator. = 31% t = 500 s k = 7.4x10-4 s-1 CHEM 3310

Make sure you can solve this equation using your calculator. The decomposition of hydrogen peroxide H2O2 (aq)  H2O (l) + ½ O2 (g) First order reaction; k = 7.4x10-4 s-1 The initial concentration of H2O2 is 2.32 M. (c) What is the time required for half of the sample to decompose? From the first order integrated equation, determine . Make sure you can solve this equation using your calculator.  = 940 s a = 1 k = 7.4x10-4 s-1 CHEM 3310

Example – Radioactive decay First Order Example – Radioactive decay Compare to the 1st order integrated equation where N0 is the number of radioactive nuclei at t=0, N is the number of radioactive nuclei at time t, and  is the decay constant. Half-life of a radioactive decay reaction is Compare to the 1st order integrated equation The decay constant, , unlike the rate constant of a chemical reaction is independent of any external influence such as pressure or temperature. CHEM 3310

First Order Example – 14C 1n + 14N → 14C + 1p 14C is created in the upper atmosphere at a steady rate. 14C decays, it forms 14N and emits a beta particle. 1n + 14N → 14C + 1p (The 14C atom has lost a neutron and gained a proton.) The half-life of the 14C decay is 5730 years. Therefore, radio-carbon dating is a method of obtaining age estimates on organic materials. Question: If we have with 1.00 mole of 14C and wait 1500 years, how much has decayed?  = 0.693 / 5730 yr-1  = 1.21x10-4 yr-1 Use this equation to calculate . , t, and N0 are known. Use this equation to calculate N. ln(N) = -t + ln(N0) N = 0.84 mole 1-N =0.16 mole decayed CHEM 3310

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

Rate Equations 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 vs. t Half-life CHEM 3310