# Chem 300 - Ch 28/#1 Todays To Do List Chemical Kinetics: Rate Law – What is it? - Experimental methods.

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Chem 300 - Ch 28/#1 Todays To Do List Chemical Kinetics: Rate Law – What is it? - Experimental methods

Rates of Reactions l How a Chem Reaction changes with time A A + B B Y Y + Z Z l At any time t: n j (t) = n j (0) ± j (t) l The change in n j (t) with time (t): d n j (t)/dt = ± j d (t)/dt Dividing by V: (1/V) d n j (t)/dt = d[ j ]/dt

d n j (t)/dt = ± j d (t)/dt l Define the rate of a reaction [v(t)]: Rate = v(t) = ± (1/ j )d[ j ]/dt l Example: 2NO + 2H 2 N 2 + 2H 2 O v(t) = - ½ d [H 2 ]/dt = + ½ d [H 2 O]/dt

The Rate Law l Relation between v(t) and the concentrations: 2NO + 2H 2 N 2 + 2H 2 O v = k [NO] 2 [H 2 ] Note lack of connection between coeff. & exponents. l In general: v(t) = k [A] m(A) [B] m(B)

v(t) = k [A] m(A) [B] m(B) l Must be determined from experimental measurements. l Usually cannot be found from balanced chemical reaction.

Rate Order l The exponents of a rate law: v(t) = k [A] m(A) [B] m(B) m(A)th order in A m(B)th order in B order overall = [m(A) + m(B)] l Example: v(t) = k[Cl 2 ] 3/2 [CO] 3/2 –order in Cl 2 1 st –order in CO Overall order = 3/2 + 1 = 2 ½

More Complex Rate Laws l Compare 2 similar reactions: H 2 (g) + I 2 (g) 2 HI(g) rate = k [H 2 ][I 2 ] H 2 (g) + Br 2 (g) 2 HBr(g) rate = k[H 2 ][Br 2 ]/(1 + k[HBr]/[Br 2 ]) Initial rate k[H 2 ][Br 2 ] These reactions occur by different processes.

Determining Rate Laws Experimentally l Method of Isolation Set up the reaction so all the reactants are in excess except one. The rate law then becomes simpler: v(t) = k[A] x [B] y k[B] y Where k = k[A] x is essentially a constant. y can be found by varying [B] and measuring change in rate (v(t)). Repeat for A with [B] in excess to find x.

Method of Initial Rates l Measure initial change in reactant conc. ( [A]) during a small time interval ( t). v - (1/ )( [A]/ t) = k[A] x [B] y Do 2 experiments: Both with same initial [A] 0 but different initial [B]: v 1 = k[A] 0 x [B] 1 y v 2 = k[A] 0 x [B] 2 y

Initial Rates contin. l Divide v 1 by v 2 : v 1 /v 2 = ([B] 1 /[B] 2 ) y l Take logs: ln(v 1 /v 2 ) = y ln([B] 1 /[B] 2 ) l Solve for y: y = ln(v 1 /v 2 ) / ln([B] 1 /[B] 2 ) l Repeat with constant [B] 0 & vary [A] to find x. l Find k by substitution into rate law.

Example of Initial Rate Exp # [A] 0 [B] 0 v(init) 10.100.104.0x10 -5 20.100.208.0x10 -5 30.200.1016.0x10 -5 l Compare expt 1 & 2 y = ln(v 2 /v 1 ) / ln([B] 2 /[B] 1 ) = ln(2)/ln(2) = 1 x = ln(v 3 /v 1 ) / ln([A] 3 /[A] 1 ) = ln(4)/ln(2) = 2 l Rate = k[A] 2 [B]

Determine rate constant l Substitute found data for one run into rate law: l Rate = k[A] 2 [B] 4.0x10 -5 = k (0.10) 2 (0.10) k = 4.0x10 -5 /0.0010 = 4.0x10 -8 l rate = (4.0x10 -8 )[A] 2 [B]

Initial Rates Contd l Alternatively, for a given [A] 0, ln v can be plotted against ln [B] and y measured from slope. l 2I(g) + Ar(g) I 2 (g) + Ar(g) Rate = k [I] 2 [Ar] 1

Next Time 1 st Order Reactions & time Different Reaction Orders Reversible Reactions

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