Presented by UGA’s Academic Resource Center 02/28/16 CHEM 1212 Exam 2 Review Presented by UGA’s Academic Resource Center 02/28/16 12 students

Chapter 14: Chemical Kinetics

Relative Rates 𝑎𝐴+𝑏𝐵→𝑐𝐶+𝑑𝐷 𝑅𝑎𝑡𝑒=− 1 𝑎 ∆ 𝐴 ∆𝑡 =− 1 𝑏 ∆ 𝐵 ∆𝑡 = 1 𝑐 ∆ 𝐶 ∆𝑡 = 1 𝑑 ∆ 𝐷 ∆𝑡 Example of instantaneous rate In terms of reactants In terms of products

Which of the following will not usually increase the rate of a chemical reaction? Crushing a solid reactant Decreasing the concentration of a reactant Increasing the temperature Liquefying a reactant B

Why should one expect an increase in temperature to increase the initial rate of reaction? Why should one expect a gaseous state reaction to happen faster than a solid state one? Is the activation energy of the gaseous reaction higher or lower than the solid state? Temperature is the measure of the average kinetic energy of the molecules. As the tempreature increases, the average energy of the particles is great enough to overcome the activation energy barrier. Gaseous molecules move freely with higher energy than molecules in solids. Statistically speaking, a higher proportion of gaseous molecules will collide with the right kinetic energy in the right orientation to react. The activation energy barrier is much lower in gases than solids.

What affects reaction rate? Temperature Increased temperature  Increased rate Surface Area Increased surface area  Increased rate Catalysts Accelerate reaction by lowering the activation energy barrier, but are not consumed Concentration As demonstrated by rate law

Rate Law Expression Order Rate Constant Units 𝑅𝑎𝑡𝑒=𝑘 𝑅𝑎𝑡𝑒=𝑘 [𝐴] 𝑚 𝑅𝑎𝑡𝑒=𝑘 [𝐴] 𝑚 [𝐵] 𝑛 Rate law (reaction orders) must be experimentally determined. Then a value of k can be calculated for a given reaction (catalyst affects k) at a given temperature. Neither time nor changing the concentration will change k. 𝑅𝑎𝑡𝑒=𝑘 [𝐴] 𝑚 [𝐵] 𝑛 [𝐶] 𝑝 k is a proportionality (rate) constant for a given reaction at a certain temperature

What’s a pseudo-first order reaction?

Initial Rate (mol/L·s) Rate of Formation of C (mol/L·s) A + 2B  3C Experiment [A] (mol/L) [B] (mol/L) Initial Rate (mol/L·s) 1 0.15 0.25 8.0 × 10-5 2 0.30 3.2 × 10-4 3 0.60 0.50 5.12 × 10-3 Experiment [A] (mol/L) [B] (mol/L) Rate of Formation of C (mol/L·s) 1 0.10 6.0 × 10-4 2 0.20 2.4 × 10-3 3 0.40 7.7 × 10-2 What’s the difference in the approach to these two determinations?

Zeroth Order Integrated Rate Law [𝑨] 𝒕 =−𝒌𝒕+ [𝑨] 𝟎

First Order Integrated Rate Law 𝐥𝐧 [𝑨] 𝒕 =−𝒌𝒕+ 𝐥𝐧 [𝑨] 𝟎

Second Order Integrated Rate Law 𝟏 [𝑨] 𝒕 =𝒌𝒕+ 𝟏 [𝑨] 𝟎

Collision Theory The reacting molecules must collide with one another. The reacting molecules must collide with sufficient energy to initiate the process of breaking and forming bonds. The molecules must collide in an orientation that can lead to rearrangement of the atoms and the formation of products.

Collision Theory The reacting molecules must collide with one another. The reacting molecules must collide with sufficient energy to initiate the process of breaking and forming bonds. The molecules must collide in an orientation that can lead to rearrangement of the atoms and the formation of products.

Reaction Coordinate Diagram

Collision Theory The reacting molecules must collide with one another. The reacting molecules must collide with sufficient energy to initiate the process of breaking and forming bonds. The molecules must collide in an orientation that can lead to rearrangement of the atoms and the formation of products.

The Arrhenius Equation The exponential term gives the fraction of molecules having sufficient energy for reaction and is a function of T. A is the frequency factor. It is related to the number of collisions and to the fraction of collisions that have the correct geometry. It is concentration dependent (it becomes smaller as the concentration of reactants become larger). It is reaction specific and temperature dependent.

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy? Increase in temperature increases the number of collisions, thereby increasing the probability of a successful collision

Mechanisms NO2 + CO  NO + CO2 2NO2  NO3 + NO (slow) NO3 + CO  NO2 + CO2 (fast) 2H2 + 2NO  N2 + 2H2O 2NO ↔ N2O2 (fast) N2O2 + H2  N2O + H2O (slow) N2O + H2  N2 + H2O (fast)

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Chapter 15: Equilibria

Equilibrium Constant Equation H2 (g) I2 (g) 2 HI(g) Initial (M) 0.0175 Change (M) -0.0138 +0.0276 Final (M) 0.0037 0.0276 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛= 𝑘 𝑓 𝐻 2 [ 𝐼 2 ] What states of matter are excluded? What are the units of K? Since these are all gases, how else could I write this expression? What equation relates Kp to Kc? 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛= 𝑘 𝑟 [𝐻𝐼] 2 𝑘 𝑓 𝑘 𝑟 = 𝐾 𝑐 = [𝐻𝐼] 2 [𝐻 2 ] [𝐼 2 ]

Magnitude of K K > 1 K < 1 Product favored Add to products, subtract from reactants K < 1 Reactant favored Add to reactants, subtract from products

Reaction Quotient Q < K Q = K Q > K Reactants  products Reaction at equilibrium Q > K Reactants  products

Stoichiometry of K C(s) + ½ O2 (g)  CO (g) 2C(s) + O2(g)  2CO(g) 𝐾 1 = [𝐶𝑂] [ 𝑂 2 ] 1/2 2C(s) + O2(g)  2CO(g) 𝐾 2 = [𝐶𝑂] 2 [ 𝑂 2 ] K2 = K12 A + B ↔ C + D Kforward = 1/Kreverse Combining equations: Add chemical equations Multiply equilibrium constants Can we have a negative value for K?

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