It is known that the reaction proceeds more rapidly in the presence of acid, provided that the hydrogen ion concentration is above 1 x 10-3 M. If the concentration of hydrogen ion is less than this, the effect of hydrogen is negligible and the rate law can be expressed by an equation of the form rate = k [H2O2]a [I-]b (Eq. 2) where k is the rate constant for the reaction, a constant that relates the speed the reaction is progressing to the concentrations of the component chemicals that are involved. Additionally, the exponents a and b are called “reaction orders”, with “a” being the order of the reaction with respect to H2O2, and “b” being the reaction order with respect to I-, for this reaction. Note: The variables a and b have specific values, that are particular to the chemical reaction that is being considered. In this exercise, you will be determining the numerical values of k, a and b. You will be also determine an estimate of the activation energy by studying the dependence of the value of k on the temperature under which the reaction is performed. General features of the method for Background on our Methods 1. The rate of the reaction has been found to be independent of hydrogen ion concentrations less than 1 x 10-3 M. The hydrogen ion concentration can be maintained at a low and constant value by carrying out this reaction in the presence of a sodium acetate-acetic acid buffer. The concentration of hydrogen ion will be maintained at about 2 x 10-5 M. The reverse reaction must be suppressed. When the concentration of iodine is allowed to build up as the reaction proceeds, the rate of a competitive reverse reaction, I2 (aq) + H2O (l) 2 I-(aq) + 2 H+(aq) + H2O2 (aq) (Eq. 3)
will increase and an equilibrium mixture will result. If the reaction in Eq. 3 occurs, your measured reaction rate will not represent the rate of the reaction described in Equation 1, but rather it will be the net difference between the forward and reverse rates (Eqs 1 and 3). Since we want to look at the rate of Eq. 1 ONLY, we need to suppress the reaction in Eq. 3 by removing the I2 as quickly as it is formed. This can be achieved by adding sodium thiosulfate, Na2S2O3, which functions chemical as 2 Na+ (aq) and S2O32- (aq) in solution.
2 Fortunately, thiosulfate, S2O32-, does not react at a measurable rate with any of the other chemical species in the solution, but it does react rapidly and completely with iodine, according to the equation I2 (aq) + 2 S2O32- (aq) 2 I- (aq) + S4O62-(aq) (Eq. 4) As long as excess thiosulfate is present in the solution, no free iodine can accumulate, and the net reaction rate will be our intended rate, for Equation 1. Notice that I- is regenerated in Eq. 4, so the concentration of I- is actually constant while there is S2O32- present. This means that the only chemical species changing in concentration, while the reaction progresses forward, is H2O2. 3. An accurate measurement of the rate at which the peroxide-iodide reaction is taking place is required for us to achieve the objective of this experiment. As described in the above points, the addition of thiosulfate facilitates our objective. Additionally, we can note that Iodine (I2) in starch solution has a blue colour. If only H2O2 and I- were present in the solution we are examining, a blue colour would appear very quickly as iodine is produced, as represented in Equation 1. However, the removal of I2 by the thiosulfate, as shown by the reaction in Equation 4, means that this solution will not turn blue until the thiosulfate is used