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Rates of Chemical Reactions, 1. The lodination of Acetone Th l the con- centrati

ID: 983784 • Letter: R

Question

Rates of Chemical Reactions, 1. The lodination of Acetone Th l the con- centrations of the reactants, the temperature, and the presence of possible catalysts. All of these factors cain he rate at which a chemical reaction occurs depends on several factors: the nature of the reaction markedly influence the observed rate of reaction. Some reactions at a given temperature are very slow inded: the oxidation of gascous hydrogen or wood at room te at at room temperature would not proceed appreciably in a century. Other reactions are essentially instantaneous: the precipitation of silver chloride when solutions containing silver ions and chloride ions are mixed and the formation of water when acidic and basic solutions are mixed are examples of extremely ra this experiment we will study a reaction that, in the vicinity of room temperature, proceeds a xed are examples of extremely rapid reactions. In relatively easily measured rate. For a given reaction, the rate typically increases with an increase in the concentration of any reactant. The relation between rate and concentration is a remarkably simple one in many cases, and for the reaction aA + bBcC the rate can usually be expressed by the equation where m and n are generally, but not always, integers: specifically 0. 1.2.r concentrations of A and B (ordinarily in moles per liter); and & is a constamt, called the rate constant of the reaction, which makes the relation quantitatively correct. The numbers m and n are called the onders of the reaction with respect to A and B. If m is 1, the reaction is said to be first order with respect to the reactant.A.If n is 2, the reaction is second order with respect to reactant B. The overall order is the sum of m and n. In this example the reaction would be third order overall. possibly 3: [A) and IB) are the The rate of a reaction is also significantly dependent on the temperature at which the reaction occurs. An increase in temperature increases the rate, an often-cited general rule being that a 10°C rise in temperature will double the rate. This rule is only approximately correct, nevertheles, it is clear that a rise in temperature As with concentration, there is a quantitative relation between reaction rate and temperature, but here the relation is somewhat more complicated. This relation is based must have a certain minimum amount of energy present at the time the reactants collide in the reaction step: this amount of energy, which is typically furnished by the kinetic energy of motion of the species present, is called the activation energy for the reaction. The equation relating the rate constant k to the absolute tempera ture T and the activation energy E, is called the Arrhenius equation: of say 100°C could change the rate of a reaction very appreciably. on the idea that to react, the reactant species In+constant where In k is the natural logarithm of k and R is the gas constant (8.31 joulesmole K for E, in joules per mole). This equation is identical in form to Equation 1 in Experiment 15. By measuring k at different tempera tures we can determine graphically the activation energy for a reaction. In this experiment we will study the kinetics of the reaction between iodine and acetone: 0 CH3-C-CHfaq) +12(aq)CH,-C-CH21(aq) +H'(aq) + r(aq)

Explanation / Answer

For second question, the innitial concentration of I is:

a) 0.0050 * (10/50) = 0.001 M

rate = 0.001 / 510 = 1.96x10-6 M/s

b) Innitial H = 1 * 10/50 = 0.2 M

Innitial acetone = 4 * (5/50) = 0.4 M

Using equation 3:

rate = k[Ac]m[I]n[H]p

rate = k(0.4)m(0.001)n(0.2)p

The unknowns are m, n, p and k.

Question 3.

a) Innitial acetone = 4 * (5/50) = 0.4 M

Innitial H = 1 * (20/50) = 0.4 M

Innitial I = 0.005 * (10/50) = 0.001 M

b) rate = 0.001 / 250 = 4x10-6 M/s

c) rate 1 = k(0.4)m(0.001)n(0.2)p

rate 2 = k(0.4)m(0.001)n(0.4)p

rate 2/rate 1 = k(0.4)m(0.001)n(0.4)p / k(0.4)m(0.001)n(0.2)p

rate 2 / rate 1 = (0.4/0.2)p

4x10-6 / 1.96x10-6 = (0.4/0.2)p

2.04 near 2 = 2p

p = 1

Hope this helps