Single Step Rate Laws. Rate Law A B where, Rate = -1 x ∆A and Rate = 1 x ∆B 1 ∆t 1 ∆t Or:...
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Transcript of Single Step Rate Laws. Rate Law A B where, Rate = -1 x ∆A and Rate = 1 x ∆B 1 ∆t 1 ∆t Or:...
![Page 1: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/1.jpg)
Single Step Rate Laws
![Page 2: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/2.jpg)
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
![Page 3: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/3.jpg)
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
![Page 4: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/4.jpg)
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
![Page 5: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/5.jpg)
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
![Page 6: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/6.jpg)
Graphs
Order can be used to predict the shape of:
Concentration vs. Time Graph
Rate vs. Concentration Graph
![Page 7: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/7.jpg)
Zero Order
• Rxn rate independent of reactant concentration
• (insert graphs from single step handout)
![Page 8: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/8.jpg)
First Order
• Reaction rate ∝ concentration• (ie. Rate decreases, concentration
decreases)
• (insert graph)
![Page 9: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/9.jpg)
Second Order
• Rate increases exponentially as concentration increases
![Page 10: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/10.jpg)
2NO N2O2
a) Write a rate lawb) Determine the order of the reactionc) Determine k if R = mol/L s when [NO] = 2.1 ∙
mol/Ld) Find the new R value if temperature is
unchanged and [NO] = 1.2 mol/Le) How does the value of R change if the [NO] is
tripled?
![Page 11: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/11.jpg)
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!
![Page 12: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/12.jpg)
a) Write a rate law
• 2NO N2O2
• Rate = k[NO]2
• (reactants only!)
![Page 13: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/13.jpg)
b) Determine the order of the reaction
• Rate = k[NO]2
• Order = 2
![Page 14: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/14.jpg)
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
Rate = k[NO]2
![Page 15: 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.](https://reader036.fdocuments.net/reader036/viewer/2022082613/5697bf8f1a28abf838c8d443/html5/thumbnails/15.jpg)
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!