Chem 104 1st Edition Lecture 19 Outline of Last Lecture I. Ideal vs. Nonideal solutionsII. Vapor pressure of a nonideal solutionIII. Colligative properties related to vapor pressure loweringIV. Freezing point depressionA. Example problems concerning freezing point depression. V. Introduction to OsmosisOutline of Current LectureI. Introduction to chemical kineticsa. Defining ratesi. Reactionii. Averageiii. InstantaneousCurrent LectureI. Reaction rate is the speed of a chemical reaction. This is measured by how fast the reaction makes products. The rate is how much a quantity changes in a given period of time. This is measured as:speed=∆ distance∆ timeChemists focus on the speed of the reaction and not the product. The rate is generally measured in terms of how much the concentration of a reactant decreases in a given period of time. A negative sign is placed in front of the reactants because the concentration decreases.rate=∆[produc t]∆ time=−∆[reactant]∆ timeAnother way to view this is:A → B−∆[A]∆ t=∆[B]∆tAs time goes on the reaction rate generally slows down because the concentration ofingredients decreases. Eventually the reaction will stop because the reactants have run out or the system has reached equilibrium. In most reactions the coefficients of abalanced equation are not the same. The change in balances has to be represented.a. Example: what is the reaction rate between CH4+2O2→CO2+2H2O?rate=−11(∆[C H4]∆ t)=−12(∆[O2]∆ t)±11(∆[C O2]∆ t)=12(∆[H2O]∆ t)II. Average rate is the change in measured concentration in any particular time period. This is linear approximation of a curve. The larger the time interval, the more the average rate deviates from the instantaneous rate. Instantaneous rate is the change in concentration at any one particular time. This is the slope of a curve. This is for more advanced chemistry.a. Example of average rate:For the reaction: H2O2+3I-+2H+→I3-+ 2H2O, [I-] changes from 1.000 M to 0.868 M in the first 10s. Calculate the average rate in the first 10s and the ∆[H+].i. First put variables into equation:− ¿I¿+¿H¿¿∆ ¿¿∆[¿¿∆t)=−12¿¿rate=− 13¿ii. Input the starting and ending M into I-:+¿H¿¿∆ ¿¿rate=−13(0.868−1.000 M10)=−12¿iii. Solve for M and multiply by -2 to both sides to cancel -1/2:+¿H¿¿∆ ¿¿rate=0.0044 M=−12¿+¿H¿¿∆ ¿¿rate=−0.0088 M=¿iv. The fact that -0.0088M is negative means the rate is
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