Single Step Rate Laws

Rate Law A  B where, Rate = -1 x ∆A and Rate = 1 x ∆B 1 ∆t 1 ∆t Or: Rate = -1 x ∆A = 1 x ∆B 1 ∆t 1 ∆t

Reaction Rate Proportional to the [concentration] of reactants raise to a power = to their coefficients in a balanced equation Rate ∝ [Reactants] coefficients Rate = k[reactants] coefficients K: – unique based on reaction – Depends on temperature

Order of Reactant = to the coefficient of a reactant in a balanced equation =the exponent of the reactant in the rate law Use these reactant coefficients to determine overall order of reaction = sum (+) of all the exponents in a rate law

Example aA + bB  cC + dD Rate ∝ [A]a[B]b Rate = k [A]a[B]b Order of A = a Order of B = b Order of Reaction = a+b

Graphs Order can be used to predict the shape of: Concentration vs. Time Graph Rate vs. Concentration Graph

Zero Order Rxn rate independent of reactant concentration (insert graphs from single step handout)

First Order Reaction rate ∝ concentration (ie. Rate decreases, concentration decreases) (insert graph)

Second Order Rate increases exponentially as concentration increases

2NO  N 2 O 2 a)Write a rate law b)Determine the order of the reaction c)Determine k if R = mol/L ∙ s when [NO] = 2.1 mol/L d)Find the new R value if temperature is unchanged and [NO] = 1.2 mol/L e)How does the value of R change if the [NO] is tripled?

Assume [NO] = 1 Then, rate = k[1] 2 = 1 x k If [NO] is tripled, then the new [NO] = 3 where Rate = k[3] 2 = 9 x k When the [NO] is tripled, rate increases by a factor of 9!

a) Write a rate law 2NO  N 2 O 2 Rate = k[NO] 2 (reactants only!)

b) Determine the order of the reaction Rate = k[NO] 2 Order = 2

c) Determine k if R = mol/L ∙ s when [NO] = 2.1 mol/L K = rate = 4.3 mol/L ∙ s = 0.98L/mol∙s [NO] 2 (2.1 mol/L) 2 Rate = k[NO] 2 Rate = k [NO] 2

d) Find the new R value if temperature is unchanged and [NO] = 1.2 mol/L Rate = k[NO] 2 = (0.98 L/mol ∙ s)(1.2mol/L) 2 = 1.4 mol/L ∙ s Don’t forget BEDMAS and order of operations!