{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

7. Reaction Kinetics AC MIALN

# 7. Reaction Kinetics AC MIALN - Reaction Kinetics and...

This preview shows pages 1–11. Sign up to view the full content.

Reaction Kinetics and Equilibrium Chapter 7

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chemical Kinetics Kinetics – study of reaction rates and their relation to the way the reaction proceeds. Reaction rate - the change in the concentration of a reactant or a product with time ( M /s). A B rate = - [A] t rate = [B] t [A] = change in concentration of A over time period t [B] = change in concentration of B over time period t Because [A] decreases with time, [A] is negative . (reactants) (products)
A B rate = - [A] t rate = [ B ] t time

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Reaction Rates and Stoichiometry 2A B 2 moles of A disappear - 1 mole of B that is formed. rate = [B] t rate = - [A] t 1 2 a A + b B c C + d D rate = - [A] t 1 a = - [B] t 1 b = [C] t 1 c = [D] t 1 d So, rate of disappearance of A is twice as fast as the rate of appearance of B
Write the rate expression for the following reaction: CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O ( g ) rate = - [CH 4 ] t = - [O 2 ] t 1 2 = [H 2 O] t 1 2 = [CO 2 ] t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers . a A + b B c C + d D Rate = k [A] x [B] y reaction is x th order in A reaction is y th order in B reaction is (x + y) th order overall
F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g ) rate = k [F 2 ] x [ClO 2 ] y rate = k [F 2 ][ClO 2 ] From exp. 1 & 2, rate 2 rate 1 k(0.10) x (0.040) y 4.8 x 10 -3 k(0.10) x (0.010) y 1.2 x 10 -3 = = 4 y = 4 y = 1 From exp. 1 & 3, rate 3 rate 1 k(0.20) x (0.010) y 2.4 x 10 -3 k(0.10) x (0.010) y 1.2 x 10 -3 = = 2 x = 2 x = 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g ) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1
Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- ( aq ) + 3I - ( aq ) 2SO 4 2- ( aq ) + I 3 - ( aq ) Experiment [S 2 O 8 2- ] [I - ] Initial Rate ( M /s) 1 0.08 0.034 2.2 x 10 -4 2 0.08 0.017 1.1 x 10 -4 3 0.16 0.017 2.2 x 10 -4

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M /s (0.08 M )(0.034 M ) = 0.08/ M s rate = k [S 2 O 8 2- ][I - ] rate = k [S 2 O 8 2- ] x [I - ] y rate 1 rate 2 k(0.08) x (0.034) y 2.2 x 10 -4 k(0.08) x (0.017) y 1.1 x 10 -4 = = From exp. 1 & 2,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}