The Effect of Concentration on Reaction Rate

ntroduction:In this experiment, we utilized the ability for the iodide ion tobecome oxidized by the persulphate ion. Our general reaction can bedescribed as:(NH4)2S2O8 + 2KI Ã I2 + (NH4)2SO4 + K2SO4 (1a)However, we know that in an aqueous solution, all of these compoundsexcept iodine will dissociate into their ionic components. Thus we canrewrite the equation in a more convenient manner:S2O82- + 2I- Ã I2 + 2SO42- (1b)It is important however to note that the NH4 and K ions are still inthe solution, they are just unreactive. In order to measure the rateof the reaction, the conventional method would be to measure thespecies in question at certain times. However, this would beinconvenient, especially for a three hour laboratory period. Since theiodide ion can be oxidized by the persulphate ion, we can use sodiumthiosulphate to be an indicator of the presence of iodine in thesolution. For this experiment, we can simply calculate the rate of thereaction by timing the amount of iodine being produced in severalruns.
The reaction between iodine and sodium persulphate can be depicted as: I2 + 2Na2S2O3 à 2NaI + Na2S4O6 (2a) Similarly, this reaction above can also be simplified due to dissociation of all the ions except for iodine and persulphate. I2 + 2S2O3 à 2I- + S4O62- (2b) An interesting property of reaction (1) is that it produces a brilliant violet colour. However, this violet colour only results in the presence of iodine, or in other words, when iodine is being produced in the reaction. If sodium thiosulphate is added to reaction (1), than as long as there are two moles of thiosulphate for every mole of iodine, the solution will be colourless because the iodine is being used up in reaction (2). However, as time passes, the thiosulphate must run out at some point, and when it does, the violet colour will appear. Timing how long it takes for the violet colour to appear will allow us to calculate the rate of the reaction. In this experiment, 5 mL was also added in order to provide a more accurate measure of the time at which the colour first appears. Starch is helpful because it forms a blue complex with free iodine. Once we have the time elapsed for each run, we can calculate the rate of the reaction by applying the equation: Rate = -âˆ?S2O8-2 / âˆ?t The change in S2O8-2 is simply half the concentration of S2O3-2 because in reaction (2), the consumption of iodine and persulphate has a 1:2 ratio. Thus, the consumption of iodine can be seen as half the consumption of persulphate (S2O3-2). After calculating the rate of the reaction, the rate constant can be found by using the equation: Rate = k [ S2O82-]m[I-] n By comparing 2 sets of data at a time from 2 different runs, the order exponents ?m? and ?n? can be calculated, and thus, we can write the rate law for the iodide-persulphate reaction. We should also expect that the expected relationships between the concentration of S2O8-2 and the rate of reaction and rate constant might not always be extremely accurate in this experiment. When dealing with ions, we must always consider the ionic strength of the ions involved. The rate of the reaction will increase as the ionic strength gets stronger, thus, we will not always see a perfect linear relationship between [S2O8-2] and the rate of the reaction and the rate constant. The purpose of this experiment was to perform many trials of the same experiment, varying only the concentrations of certain ionic compounds in order to determine the affect of concentration on the rate of a reaction. By varying the concentration of different compounds for each run while keeping other factors constant, we were able to obtain experimental data that would give us a relationship between concentration and reaction rate. Experimental Procedure: Refer to Experiment 3, pages 3-1 to 3-4 within First Year Chemistry: Chem 123L Laboratory Manual Experimental Observations: Table 1 ? Data Sheet Add to Erlenmeyer Add to Beaker Run # 0.1 M (NH4)S2O8 (mL) 0.1 M (NH4)2SO4 (mL) 0.2 M KI (mL) 0.2 M KNO3 (mL) 0.01M Na2S2O3 (mL) H2O (mL) 0.2% Starch (mL) Elapsed Time (s) 1 20 0 20 0 10 0 5 113.5 2 10 10 20 0 10 0 5 218.2 3 5 15 20 0 10 0 5 476.3 4 20 0 10 10 10 0 5 228.2 5 20 0 5 15 10 0 5 431.9 6 10 0 20 0 10 10 5 183.4 7 20 0 5 0 10 15 5 574.6 Results and Calculations: 1. Sample Calculation: [S2O8-2], [I-], [S2O3-2], -âˆ?S2O8-2 =? mol L-1 Run #1: [S2O8-2] = 20 mL (NH4)S2O8 x 0.1 mol (NH4)S2O8 = 3.64 × 10-2 mol L-1 55 mL total 1 L (NH4)S2O8 [I-] = 20 mL KI x 0.2 mol KI = 7.27 × 10-2 mol L-1 55 mL total 1 L KI [S2O3-2] = 10 mL Na2S2O3 x 0.01 mol Na2S2O3 = 1.82 × 10-3 mol L-1 55 mL total 1 L Na2S2O3 -âˆ?S2O8-2 = 1/2 [S2O3-2] = -9.09 × 10-4 mol L-1 2. [image] [image] From plots 1 &2: Since; âˆ?S2O8-2 = k [S2O8-2]m[I] n âˆ?t and since âˆ?S2O8-2, k and [I]n are constant in runs 1, 2 and 3 (referring to plot 1), we can rewrite the equation to yield: 1 = B [S2O8-2]m where B = k [I-]n âˆ?t âˆ?S2O8-2 Thus we can find the exponent the order ?m? with respect to S2O8-2 by writing the equation of the line in the form of y = mx + b where m= slope of plot 1= exponent i.e. ? log âˆ?t = log B + m log [S2O8-2] Similarly, we can find the order ?n? with respect to I- by using the equation 1 = A [I-]n where A = k [S2O8-2]m âˆ?t âˆ?S2O8-2 and by rewriting the equation of the line in the form of y = nx + b where n= slope of plot 2 = exponent i.e. ? log âˆ?t = log A + n log [I-] slope 1: -log âˆ?t vs. log [S2O8-2] m= rise = -2.678 ? (-2.055) = -0.623 = 1 run -2.041 ? (-1.439) -0.602 slope 2: log âˆ?t vs. log [I] n= rise = -2.635 ? (-2.055) = -0.58 = 1 run -1.74 ? (-1.138) -0.602 3. Sample Calculations: Rate of Reaction for run #1: Rate = -âˆ?S2O8-2 / âˆ?t = -9.09 × 10-4 M / 113.5 s = 8.01 × 10-6 M s-1 Rate constant (k) for run #1: k = Rate / ( [S2O8-2]m [I-]n ) k = 8.01 × 10-6 M s-1 / {(3.64 × 10-2 M)1(7.27 × 10-2 M)1} k = 3.03 × 10-3 s-1 Ionic Strength (Â?) for run #1: Â? = 0.5 Σ CiZi2 Â? = 0.5{([NH4]x(1)2) + ([S2O8]x(-2)2) + ([K]x(1)2) + ([I]x(-1)2) + ([Na]x(+1)2) + ([S2O3]x(-2)2)} Â? = 0.187 mol L-1 Table 2 ? Calculations Summary Table Run # [S2O8-2] (M) [I-] (M) [S2O3-2] (M) -âˆ?S2O8-2 (M) âˆ?t (s) Rate (M s-1) Rate Constant, k (s-1) Ionic Strength (M) 1 3.64 × 10-2 7.27 × 10-2 1.82 × 10-3 -9.09 × 10-4 114 -8.01 × 10-6 3.03 × 10-3 0.187 2 1.82 × 10-2 7.27 × 10-2 1.82 × 10-3 -9.09 × 10-4 218 -4.16 × 10-6 3.15 × 10-3 0.187 3 9.09 × 10-3 7.27 × 10-2 1.82 × 10-3 -9.09 × 10-4 476 -1.91 × 10-6 2.89 × 10-3 0.187 4 3.64 × 10-2 3.64 × 10-2 1.82 × 10-3 -9.09 × 10-4 228 -3.98 × 10-6 3.01 × 10-3 0.187 5 3.64 × 10-2 1.82 × 10-2 1.82 × 10-3 -9.09 × 10-4 432 -2.10 × 10-6 3.18 × 10-3 0.187 6 1.82 × 10-2 7.27 × 10-2 1.82 × 10-3 -9.09 × 10-4 183 -4.96 × 10-6 3.75 × 10-3 0.133 7 3.64 × 10-2 1.82 × 10-2 1.82 × 10-3 -9.09 × 10-4 575 -1.58 × 10-6 2.39 × 10-3 0.133 Questions: 1. The average rate constant for this reaction for runs 1 to 5 can be calculated using the equation: Avg. rate constant = (k1 + k2 + k3 + k4 + k5)/ 5 = 3.05 × 10-3 s-1 Therefore, the average rate constant is approximately 3.05 × 10-3 s-1. 2. The k value for run #2 is close to the k value of run #6 ? it has a deviation of approximately 16%. However, the k value for run #5 is still similar to the k value for run #7, but it has a higher deviation than the first comparison ? it has a deviation of approximately 25%. We can say that all the k values for all 7 runs are similar to one another since the difference between each trial is not much. However, we can account for run #5 and run #7 having a larger deviation between their k values because it has a larger difference in volume. Runs #2 and #6 only differ with 10 mL of 0.1 M (NH4)2SO4. An extra addition of 10 mL of (NH4)2SO4 to the reaction will cause a slight decrease in the rate of the entire reaction as we can see in Table 1. Run #2 has a faster rate of reaction than run #6 because when we added more (NH4)2SO4, it hindered the product of more iodine, which will in turn take longer to react completely with the sodium thiosulphate, our iodine indicator. Since we calculated our rate of reaction based on the disappearance of the sodium thiosulphate colour, if there is hindering of the production of iodine, it will result in a slower rate of reaction. Runs #5 and 7 have k values that have a larger deviation because unlike runs #2 and 6, they differ by 15 mL of 0.2 M KNO3. Run
had the extra 15 mL of KNO3 and resulted in a faster reaction rate
because it pushes the reaction in the direction that produces more
iodine, and this in turn will result in a faster rate of reaction for
the same reasons above.
3. It is very important when reporting a rate constant for a reaction that the ionic strength and temperature be stated. Ionic strength and temperature of the solution and/or surroundings can have immense effects on the rate of the reaction. This can affect the rate constant heavily because these are two factors that affect the rate of reaction, which is linearly proportional to the rate constant. For example, if we were to raise the temperature of the solution, the rate of the reaction would increase. Ionic strength can also have a great influence on the rate constant and the rate of reaction; at low concentrations of reactants, the majority of effects on the rate of reaction can be from ionic strength, disregarding the actual chemical identity of the ion. 4. If the S2O8-2 concentration were doubled, the rate of reaction and rate constant would both be affected. The rate constant depends on the concentration of S2O8-2; they have a linear relationship. That means that if the concentration of S2O8-2 is doubled, then ? assuming everything else remains constant ? the rate constant will double as well. It is clear that if the rate constant changes than the rate of reaction should change as well. If every other component is constant and the only variation occurs with the concentration of S2O8-2 concentration, the rate of reaction is linearly proportional to the S2O8-2 concentration. That means that if the concentration doubles than the rate of the reaction should double, more or less. This can be quite clear when comparing runs #1 and 2. Discussion: By measuring the amount of iodine consumed after a certain amount of time elapsed, we were able to calculate the rate of the overall reaction. Knowing that the concentration of S2O8-2 was linearly proportional to the rate constant, we can see in the results that the relationship is very close to being linear, but there is some variation. This can be explained by the ionic strength factor. Since the reaction involves 2 negatively charged ions, the rate is greatly dependant on ionic strength; the greater the ionic strength, the faster the rate of reaction. In order to optimize the conditions for this experiment, there must be an addition of an electrolyte such as (NH4)2SO4 or KNO3 when the reactant ions are being lowered to keep the ionic strength constant. For example in runs #1 and 6, the concentration of the persulphate ion has decreased by a half in run #6 and the iodide ion is constant; this should give the rate constant in run #6 a decrease by a half. However we can see that this is not the case. Since the ionic strength has also decreased, it has some effect on the resulting rate constant and therefore skews the results a bit. The rest of the results seem to agree with the logical way the experiment should have occurred. For example, the runs with the longer elapsed times had the slower reaction rates and vice versa with the runs with the shorter elapsed times. This makes sense due to the linear relationship between reaction rate and time. Some sources of error in this experiment may have been a mistake in mixing certain reactants, or inaccuracy with measuring volume of the solutions. It was more likely that there was inaccurate measuring of the solutions because it was quite difficult to always use the Mohr and transfer pipettes precisely. Conclusions: The purpose of this experiment was to determine how concentration of a certain reactant in a reaction can affect the rate of the entire reaction. The experiment was overall a success because we could see that when we varied the concentrations of certain compounds, the reaction rate was affected accordingly. Overall we know that the rate of the reaction is linearly proportional to the concentration of your reactant. However, if your reaction can exist in equilibrium and you increase the concentration of a product, the reaction will favour in the left direction, and if you are measuring rate of product formation, this will result in a decrease in reaction rate.

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