Rate Equations and Order of Reactions 14 Rate Equations and Order of Reactions Increasing concentration of reactants can increase the rate of reaction. Is there any mathematical relationship between rate of reaction and concentration of reactants? YES!!!!!!

2NO(g) + 2H2(g)  N2(g) + 2H2O (l) For the reaction: 2NO(g) + 2H2(g)  N2(g) + 2H2O (l) Initial concentration / mol dm-3 Initial rate / mol dm-3 s-1 NO H2 0.0250 0.100 2.4  10-6 0.050 1.2  10-6 0.0125 0.6  10-6 Rate a [H2(g)] when [NO(g)] is constant Rate a [NO2(g)]2 when [H2(g)] is constant Rate a [NO(g)]2 [H2(g)] Rate = k[NO(g)]2 [H2(g)]

Rate Equations and Order of Reactions 14.1 Rate Equations and Order of Reactions

Rate Equation In general, for the reaction: mA + nB  products Rate = 14.1 Rate Equations and Order of Reactions (SB p.25 NB p.8) Rate Equation In general, for the reaction: mA + nB  products Rate = = k [A]x [B]y m n (a rate equation) k = rate constant x = order of reaction with respect to reactant A y = order of reaction with respect to reactant B Overall order of reaction = x + y

Rate Equation mA + nB  products Rate = k [A]x [B]y 14.1 Rate Equations and Order of Reactions (SB p.25) Rate Equation mA + nB  products Rate = k [A]x [B]y x and y are determined experimentally. x and y may NOT equal to m and n. x and y may not be whole numbers!!! Rate of reaction is independent of products’ concentration

Order of Reactions Usually integers (0, 1, 2, … …) If x = 1 14.1 Rate Equations and Order of Reactions (SB p.26) Order of Reactions Do Class Examples on p. 8 Usually integers (0, 1, 2, … …) If x = 1  first order with respect to reactant A If y = 2  second order with respect to reactant B The overall reaction is a third order reaction.

Rate Equation Rate = k [A]x [B]y 14.1 Rate Equations and Order of Reactions (SB p.25 NB p. 8) Rate Equation Rate = k [A]x [B]y log rate = log k + x log[A] + y log[B] If B is kept large excess, its change in concentration will be negligible when compared with A.  log[B] ~ constant  log rate = c’ + x log[A]  when is log rate plotted against log[A]  straight line with slope = x

Zeroth, First and Second Order Reactions 14.2 Zeroth, First and Second Order Reactions

For the following reaction: A  product [A] mol dm-3 Time (s) 1.00 0.80 10 0.70 20 0.65 30 0.53 40 : How can we find out the order of reaction with respects to [A]?

Zeroth Order Reactions 14.2 Zeroth, First and Second Order Reactions (SB p.27 NB p.10) Zeroth Order Reactions A  products Rate = k [A]0 = k e.g. CH3COCH3(aq) + I2(aq) + H+(aq)  CH3COCH2I(aq) + I-(aq) + 2H+(aq) Rate = k[CH3COCH3(aq)][H+(aq)][I2(aq)]0 The rate of reaction is independent of [I2]!!! This means if we keep [CH3COCH3(aq)] and [H+(aq)] constant then measure the initial rate with various [I2(aq)], we will get the same initial rate.

Zeroth Order Reactions 14.2 Zeroth, First and Second Order Reactions (SB p.27) Zeroth Order Reactions [I2(aq)]

Zeroth Order Reactions WHY??? 14.2 Zeroth, First and Second Order Reactions (SB p.27) Zeroth Order Reactions WHY??? The reaction is a multi-step reaction. Only the slowest step affect the overall reaction rate. A B C D Iodine is not involved in the slowest step of the reaction.

A Plot of [A] against time will give a straight line 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.37) Zeroth Order Reaction A Plot of [A] against time will give a straight line A  product Rate equation is: [A] = -kt + [A]0

First Order Reaction e.g Decomposition of H2O2(aq) to H2O and O2 14.2 Zeroth, First and Second Order Reactions (SB p.28) First Order Reaction e.g Decomposition of H2O2(aq) to H2O and O2 2H2O2(aq)  2H2O(l) + O2(g) Rate = k [H2O2(aq)] Though the stoichiometric coefficient of H2O2(aq) is 2 Order of reaction with respect to H2O2(aq) is 1 (determined experimentally)

A Plot of ln [A] against time will give a straight line 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.37 NB p. 9)) First Order Reaction A Plot of ln [A] against time will give a straight line A  product Rate equation is: In A = -k1t + In [A]0

Half-life of First Order reaction 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.38) Half-life of First Order reaction The time taken for half of the reactant to be converted to the product is known as the half-life of the reaction. After time t, [A] = [A]0 t is called the half life of the reaction.

Half-life of First Order Reaction 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.38) Half-life of First Order Reaction ln[A] = -k1t + ln[A]0 k1t = ln[A]0 - ln[A] If t = half-life (t0.5), then [A]0 = 2[A]: k1t0.5 = ln 2 t0.5 = constant

Half-life of a First Order Reaction 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39) Half-life of a First Order Reaction For a first order reaction: Half-life is a constant Do Class Examples 1, 3 on p.10

Second Order Reaction A  product Rate equation is: 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40 NB p.11) Second Order Reaction A plot of 1/[A] vs time will give a straight line A  product Rate equation is: Integrating the above rate equation, obtain:

Summary Plot a second graph of vs time Do Q. 5 on p. 77 Do Q. 14 on p. 80 Q. 13 (more difficult) Do Q. 3 on NB p.17 Q. 2 on NB p. 21 Summary Steps to determine the order of reaction Plot a concentration-time graph  a straight line  zeroth order  a curve with constant half life  first order Confirmation: ln[A] vs time  straight line Plot a second graph of vs time  a straight line  second order

Determination of Simple Rate Equations by Graphical Method 14.4 – 14.5 Determination of Simple Rate Equations by Graphical Method Notes p. 18

log (rate)-log (concentration) Graph 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.32) log (rate)-log (concentration) Graph rate = k[A]n log (rate) = n log [A] + log k Plotting log (rate) against log [A], a straight line is obtained for any orders Slope of straight line = Order of reaction (n) y-intercept = log k Do Q. 1 on NB p. 16

We need to plot two graphs: (1) [A] vs time to find the rate at different [A] (by slope to the curve) (2) log (rate) vs log[A] to find the order and k

14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) Check Point 14-4 Decide which curve in the following graph corresponds to (i) a zeroth order reaction; (ii) a first order reaction. (i) (3) (ii) (2)

Isolation technique When two or more substances react: 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.33) Do Q. 1, 3 on p. 27 Do Q 13 on p.80 Isolation technique When two or more substances react: A + B  products Order of reaction with respect to reactant A can be found by keeping concentration of B constant (by using much excess B – isolation technique) Rate = k[A]x[B]y  Rate = k’ [A]x

The END

Q.13 on p. 80 Products (not reactants) t(s) v(cm3) 5 33 10 56 15 73 25 92 3600 110 5400 ln (a) vol. of N2 (a) in reactant/cm3 110 77 54 37 18