Unit 10 Kinetics
The rate law for a reaction describes the dependence of the initial rate of a reaction on the concentrations of its reactants. It includes the Arrhenius constant, k, which takes into account the activation energy for the reaction and the temperature at which the reaction occurs. The rate of a reaction is described in terms of the rate of appearance of a product or the rate of disappearance of a reactant. The rate law for a reaction cannot be determined from a balanced equation. It must be determined from experimental data, which is presented in table form.
Determining rates
The Rate Law Let s look at a rate law Rate = k[a] x [B] y [C] z -The unit on Rate will be concentration/time -The unit on [A], [B], or [C] will be moles/l -x is the order of reactant A, y is the order of reactant B, and z is the order of reactant C - x + y + z = the overall order of reaction - k is the rate constant and its unit must be determined
Determining order of reactant If the concentration of a reactant doubles and the rate remains the same If the concentration of a reactant doubles and the rate doubles If the concentration of a reactant doubles the rate quadruples Zero order in that reactant First order in that reactant Second order in that reactant
NH 4+ + NO 2- N 2 + H 2 O Exp Initial [NH 4+ ] Initial [NO 2- ] Initial Rate (M/s) 1 0.100 M 0.0050 M 1.35 x 10-7 2 0.100 M 0.010 M 2.70 x 10-7 3 0.200 M 0.010 M 5.40 x 10-7 Exp 1&2 [NH 4+ ] same; [NO 2- ] x 2; rate x 2; therefore 1 st order [NO 2- ] Exp 2&3 [NH 4+ ] x 2; [NO 2- ] same; rate x 2; therefore 1 st order [NH 4+ ] Overall 2 nd order reaction Rate = k[nh 4+ ][NO 2- ] Using Exp 2 info 2.70 x 10-7 M/sec = k(.100 M)(.100 M) k = 2.70 x 10-5 M -1 sec -1
BrO 3- + 5Br - + 6H + 3Br 2 + 3H 2 O Exp [BrO 3- ] [Br - ] [H + ] Initial Rate M/s 1 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.20 0.20 0.10 3.2 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3 Exp 1&2 [BrO 3- ] x 2; [Br - ] same; [H + ] same; rate x 2; therefore 1 st order [BrO 3- ] Exp 2&3 [BrO 3- ] same; [Br - ] x 2; [H + ] same; rate x 2; therefore 1 st order [Br - ] Exp 1&4 [BrO 3- ] same; [Br - ] same; [H + ] x 2; rate x 4; therefore 2 nd order [H + ] Overall 4 th order reaction Rate = k[bro 3- ][Br - ][H + ] 2 Using Exp 1 info 8.0 x 10-4 M/sec = k(.10 M)(.10 M)(.10 M) 2 k = 2.70 x 10-5 M -1 sec -1
Let s look at rate constants Order of Reaction Possible Rate Law Units on k Zero order Rate = k[a] 0 M time -1 First order Rate = k[a] time -1 Second order Rate = k[a] 2 M -1 time -1 Third order Rate = k[a] 3 M -2 time -1
Another way to determine rate order You can determine the order of a reactant by graphing concentration data vs time. If [A] vs time gives a straight line, then the order of reactant is zero order. If ln[a] vs time gives a straight line, then the order of reactant is first order. If 1/[A] vs time gives a straight line, then the order of reactant is second order.
Zero order rate laws The rate of a zero-order reaction does not depend on the concentration of reactants at all, so the rate of a zero-order reaction will always be the same at a given temperature. Rate = k The graph of [A] vs time will be a straight line with a slop equal to k.
For a zero order reaction, as shown in the following figure, the plot of [A] versus time is a straight line with k = - slope of the line. Other graphs are curved for a zero order reaction.
Rate = k[a] First order rate law As the [A] is depleted over time, the rate of reaction will decrease with a characteristic half-life. This is the same curve used for nuclear decay, which is also a first order process. It is a plot of ln[a] vs time Half-life =.693/k Therefore, k =.693/half-life
For a first order reaction, as shown in the following figure, the plot of the logrithm of [A] versus time is a straight line with k = - slope of the line. Other graphs are curved for a first order reaction.
Second order rate laws Rate = k[a] 2 The rate of this second order reaction depends on the concentration of a single reactant raised to the second power. Plotting 1/[A] vs time creates a linear graph Slope of the line is given by k The y-intercept is given by 1/[A] 0
For a second order reaction, as shown in the following figure, the plot of 1/[A] versus time is a straight line with k = slope of the line. Other graphs are curved for a second order reaction.
Summary
Collision theory According to the collision theory, chemical reactions occur because reactants are constantly moving around and colliding with one another. When reactants collide with sufficient energy (activation energy, E a ), a reaction occurs. Reaction rate increases with increasing concentration of reactants because if there are more reactant molecules moving around in a given volume, then more collisions will occur. Reaction rate increases with increasing temperature because increasing temperature means that the molecules are moving faster, which means that the molecules have greater average kinetic energy. The higher the temperature, the greater the number of reactant molecules colliding with each other with enough E a to cause a reaction.
The Boltzmann distribution diagram below is often used to show that increasing temperature increases the fraction of reactant molecules above the activation energy.
The Rate Constant & Temperature The rate constant will increase when temperature in increased. The relationship between temperature and the rate constant is given by the Arrhenius equation, k -E a RT = Ae This equation is generally rewritten using natural logarithms. Ea 1 lnk= - + lna R T k = rate constant E a = activation energy R = gas constant, 8.31 J/kmol T = Kelvin temperature A = a constant that takes into acct collision frequency & orientation When lnk is graphed vs 1/T, it makes a straight line with a slope of E a /R.
Kinetics and Equilibrium There is a relationship between the rate constants for the forward and reverse directions of a particular reaction and the equilibrium constant for that reaction K eq = k k f r K eq = the equilibrium constant k f = the rate constant for the forward reaction k r = the rate constant for the reverse reaction
Catalysts A catalyst increases the rate of a chemical reaction without being consumed in the process A catalyst does not appear in the balanced equation. A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy.
Reaction Mechanisms Many chemical reactions are not one-step processes Rather, the balanced equation is the sum of a series of simple steps. For instance, the hypothetical reaction 2 A + 2 B C + D Rate = k[a] 2 [B] could take place by the following three-step mechanism: I. A + A X (fast) II. X + B C + Y (slow) III. Y + B D (fast) Species X and Y are called intermediates because they appear in the mechanism, but they cancel out of the balanced equation.
Reaction Mechanism, cont. Let s show that the mechanism is consistent with the balanced equation. I. A + A X (fast) II. X + B C + Y (slow) III. Y + B D (fast) A + A + X + B + Y + B X + C + Y + D Cancel species that appear on both sides and you get 2 A + 2 B C + D (for a mechanism to be plausible, it must add up to be the balanced equation) Step II is called the rate determining step.
The rate law for the reaction Cl 2 + H 2 S S + 2HCl is found to be Rate = k[cl 2 ][H 2 S]. Which of the following mechanisms are consistent with the rate law? (a) Cl 2 Cl + + Cl - (slow) Cl - + H 2 S HCl + HS - (fast) Cl + + HS - HCl + S (fast) (b) Cl 2 + H 2 S HCl + Cl + + HS - Cl + + HS - HCl + S (fast) (slow) agrees with rate law (c) Cl 2 Cl + Cl (fast, equilibrium) Cl + H 2 S HCl + HS (fast, equilibrium) HS + Cl HCl + S (slow)