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# The determination of a rate equation.

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Introduction

Planning: The determination of a rate equation Aim: Determine how the concentration or volume of each component affects the rate equation of the reaction between sodium thiosulphate Na2S2O3(aq) and hydrochloric acid HCl(aq), using a graphical method. Theory: The rate of a reaction can be expressed in terms of the rate of decrease in concentration of a reactant or the rate of increase in concentration of a product. It is affected by the following factors: concentration of reactants, pressure, temperature, physical state of reactants and catalysts. The most common method used to calculate the rate of reaction is to measure the change in concentration of the reactant(s) per second. The rate of a reaction may be represented by a mathematical equation related to the chemical equation for a reaction. Rate equation has the form rate = k [A] x [B] y which shows how the rate of a chemical reaction depends on the concentration of reactants (A and B) and the rate constant k. The rate equation states the relationship between the rate of reaction and the concentration of each reactant: rate ? [reactant]n which can also be written as: rate = k[reactant]n where k is the rate constant of the reaction, which means the change in concentration per unit time of reactant or product in a reaction in which all the reactants are at unit concentration. ...read more.

Middle

the chemicals The volume of the 0.4M solution of S2O32- for dilution to 0.200 mol dm-3 0.4 x V/1000 = 0.2 x 250/1000 V = 125 cm3 The volume of the 0.4M solution of S2O32- for dilution to 0.104 mol dm-3 0.4 x V/1000 = 0.1 x 250/1000 V = 65 cm3 The volume of the 0.4M solution of S2O32- for dilution to 0.240 mol dm-3 0.4 x V/1000 = 0.24 x 250/1000 V = 150 cm3 The volume of the 0.4M solution of S2O32- for dilution to 0.304 mol dm-3 0.4 x V/1000 = 0.304 x 250/1000 V = 190 cm3 The volume of the 0.2 solution of HCl for dilution to 1.60 mol dm-3 2.0 x V/1000 = 1.5 x 50/1000 V = 40 cm3 The volume of the 2 solution of HCl for dilution to 0.60 mol dm-3 2.0 x V/1000 = 1.5 x 50/1000 V = 15 cm3 The volume of the 2 solution of HCl for dilution to 1.32 mol dm-3 2.0 x V/1000 = 1.4 x 50/1000 V = 33 cm3 The volume of the 2 solution of HCl for dilution to 1.00 mol dm-3 2.0 x V/1000 = 1.0 x 50/1000 V = 25 cm3 Concentration (mol dm-3) 0.60 1.00 1.32 1.60 2.00 Volume of HCl (cm3) 15 25 33 40 50 Volume of Distilled water (cm3) ...read more.

Conclusion

3. Deduce the order of reaction with respect to each reactant from the shape of the graph. * Zero order The graph would be a horizontal line, which means the respective reactant doesn't have any effect on the reaction rate. rate ? [reactant]0 * First order A straight line passing through the origin will be obtained, i.e. the rate is directly proportional to the concentration of the reactant. The rate constant, k can be deduced from the slope of the line (rate/[reactant]). rate ? [reactant]1 * Second order The rate of reaction is directly proportional to the square of the reactant's concentration (rate ? [reactant]2), i.e. when a square amount of volume of reactant is used, the rate would be double. In this case, we will get a curve with an increasing gradient. However, if we plot a graph of 1/t against the square of volume used, we can get a straight-line graph, which go through the origin. rate ? [reactant]2 1/t /10-2 s-1 2nd 1st 0 Volume 4. Write out the rate equation of reaction by combining the reaction rates. rate ? [S2O32-(aq)]n rate ? [H+(aq)]m where n and m are the order of reaction of S2O32-(aq) and H+(aq) respectively. The overall reaction rate is: rate ? [S2O32-(aq)]n [H+(aq)]m So, the rate equation is: rate = k [S2O32-(aq)]n [H+(aq)]m where k is the rate constant and (n + m) is the overall order of the reaction. ...read more.

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