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1. Give a rate law for a zeroth-order reaction. What sort of plot needs to be made to give a straight line? What are the units of k?

*Rate = k*

*For a zeroth-order reaction, the rate is constant. The rate is not dependent on concentration at all. A plot of k versus time gives a straight line. The units of k are M/s.*

2. Give a rate law for a first-order reaction. What sort of plot needs to be made to give a straight line? What are the units of k?

*Rate = k[A] ^{1}*

*A first-order reaction has a rate proportional to the concentration of one of the reactants. A plot of ln[A] versus t gives a straight line. The units of k are s ^{-1}.*

3. Give a rate law for a second-order reaction? What sort of plot needs to be made to give a straight line? What are the units of k?

*Rate = k[A][B] or Rate = k[A] ^{2}*

*A plot of 1/[A] versus t gives a straight line. The units of k are M ^{-1}s^{-1}.*

4. How does one go about experimentally determining a rate law?

*If a reaction involves only one reactant, the rate law can be determined by measuring the initial rate of the reaction as a function of the reactant's concentration. For example, if the rate doubles when the concentration of the reactant doubles, the reaction is first order in the reactant. If the rate quadruples when the concentration doubles, the reaction is second order in the reactant.*

*For a reaction involving more than one reactant, one can find the rate law by measuring the dependence of the reaction rate on the concentration of each reactant, one at a time. The concentration of all but one reactant can be held constant and the rate of the reaction can be recorded. Any changes in the rate must be due only to changes in the reactant whose concentration was changed. The same procedure is applied to the next reactant until the order of all reactants is determined. Ta da!*

5. 2NO + 2H_{2} → N_{2} + 2H_{2}O

Experiment | [NO] (M) | [H_{2}] (M) | Initial Rate (M/s) |
---|---|---|---|

1 | 5.0 x 10^{-3} | 2.0 x 10^{-3} | 1.3 x 10^{-5} |

2 | 10.0 x 10^{-3} | 2.0 x 10^{-3} | 5.0 x 10^{-5} |

3 | 10.0 x 10^{-3} | 4.0 x 10^{-3} | 10.0 x 10^{-5} |

Using the above data, determine 1) the rate law, 2) the rate constant, and 3) the rate of the reaction when [NO] = 12.0 × 10^{-3 }M and [H_{2}] = 6.0 × 10^{-3 }M.

*Explanation**Rate = k[NO] ^{x}[H_{2}]^{y}*

*Rate _{2} / Rate_{1} = 5.0 × 10^{-5 }M/s / 1.3 × 10^{-5} M/s = ~4*

4 = k(10.0 × 10^{-3})^{x}/ k(5.0 × 10^{-3})^{x}

*Cross out what cancels*

*4 = (10.0 × 10 ^{-3})^{x}/(5.0 × 10^{-3})^{x }= 2^{x} = 4 *

*x = 2*

*Rate _{3} / Rate_{2} = 10.0 × 10^{-5 }/ 5.0 × 10^{-5} = 2*

*2 = k (4.0 x 10 ^{-3})^{y} / (2.0 × 10^{-3})^{y}*

*Cross out what cancels*

*2 = (4.0 × 10 ^{-3})^{y }/ (2.0 × 10^{-3})^{y} = 2^{y}*

*y = 1*

**rate = k[NO]^{2}[H_{2}]**

*rearranging the law and we get*

*k* = rate/ [NO]^{2}[H_{2}]

*using the data from experiment 2 we get*

*k = 5.0 × 10 ^{-5 }/ (10.0 × 10^{-3})^{2} (2.0 × 10^{-3}) = 2.5 × 10^{2 }M^{2}s*

*Using the known rate constant and concentrations of NO and H _{2}, we write*

*Rate = (2.5 × 10 ^{2})(12.0 × 10^{-3})^{2}(6.0 × 10^{-3}) = 2.2 × 10^{-4 }M/s*

6. What is a half-life? How do we calculate what it is for a first-order reaction?

*The half-life of a reaction, t _{1/2}, is the time required for the concentration of a reactant to decrease to half of its initial concentration. Rearranging the first-order rate law gives us:*

*t = 1/k ln [A] _{0} / [A]_{t}*

*By the definition of a half-life we know t = t _{1/2 }[A] = [A]_{0}/2 so*

*t _{1/2 }= 1/k ln [A] / [A]_{0}/2*

*t*_{1/2}= 1/*k* ln 2 = 0.693 / *k*

*This equation tells us that the half-life of a first-order reaction is independent of the initial concentration of the reactant.*

7. Consider the reaction of A → 2B. This reaction has a rate constant of 5.36 × 10^{-4 }s^{-1}. Calculate the half-life of this reaction in minutes.

*t _{1/2}=0.693/k*

*= 0.693 / 5.36 × 10 ^{-4 }= 1.29 × 10^{3 }s × 1 min/ 60 s = 21.5 minutes*

8. What is the collision theory? How is temperature related?

*Collision theory qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The theory states that molecules must collide before they can react. In order to initiate a reaction these collisions must be sufficiently energetic (kinetic energy) to bring about bond disruption. Increasing the temperature makes molecules move faster and collide more vigorously greatly increasing the likelihood of bond rearrangements.*

9. Consider 2A + B ⇌ 2C Δ*H* = -197 kJ/mol

*How can one shift the equilibrium to the right?*

*Add a catalyst, add more A, add more B, or increase the temperature*

10. What is Le Chatelier's Principle?

*Le Chatelier's Principle states that if any change is imposed on a system that is in equilibrium, then the system tends to adjust to a new equilibrium to counteract the change.*