Hi, i completed a kinetic experiment That involves adding 20ml of persulfate and

Question

Hi, i completed a kinetic experiment That involves adding 20ml of persulfate and 10 ml Iodine with 10ml thiosulfate ... A reaction ( Rxn) occurs in the 50 ml mixture and all the thiosulfate was consumed in a given time ( 97sec) indicated by the starch turning blue ... The lab instructions asks me to calculate the reaction rate ( initial rate) but i have trouble understanding what the instruction is asking me to do .. Can someone show me ( visually) how to perform this reaction rate calculation ? Thank You le08rd the temperature of the reaction solution. Data Analysis and Calculations. cat Calculating the Reaction Rate. Remember, the starch turns blue when all the thiosulfate has been consumed. To calculate the reaction rate (the initial rate) for each trial: 1. Calculate the initial moles ofthiosulfate S measured time interval. 2. Based on the molesof thiosulfate consumed use the stoichiometric relations from reactions #1 and #2 on the first page) to figure out how many moles ofpersulfate (S2082) are consumed in reaction #1 during the same time interval. This is the change in moles of persulfate during the reaction. 3. Since the rate of the reaction is defined as the negative of the change in molarity of persulfate per time, calculate the reaction rate by dividing the change in moles of persulfate by the volume of the reaction solution and dividing by the time (in seconds). The formula is: initial moles of thiosulfate (S.0 803 s). This is the amount consumed in reaction #2 in the rate--afs,og]-moles ofs,of consumed time (in seconds, The formula is: 0.005 M of ,0 initial,,-)0.000M of-272 Tme9 rate»-|s,08-= moles of S202-consumed initial seC At Lof reaction solution time molart Note: In this calculation we are assuming that the rate stays essentially constant during the measured time. This is a reasonable assumption since we are only running the reaction for a short period of time, so the [S2O and the [r] don't change very much compared to their initial values

Explanation / Answer

Write down the two reactions:

S2O82- (aq) + 2 I- (aq) -------> 2 SO42- (aq) + I2 (aq) ……..(R1)

The molar ratio of S2O82-:I2 is 1:1.

Again, the second reaction is

2 S2O32- (aq) + I2 (aq) -------> S4O62- (aq) + 2 I- (aq) ……..(R2)

The molar ratio of S2O32-:I2 is 2:1.

Therefore combining (R1) and (R2), we can say that the molar ration of S2O82-:S2O32- is 1:2.

Now we shall calculate the moles of the two reactants.

The total volume of the solution = (20 + 10 + 10) mL = 40 mL; you have however written 50 mL as the total volume and I believe this takes into account the addition of starch and possibly some dilution by water. I will show you the calculation using 50 mL as the volume of the solution.

Initial molarity of S2O32- = 0.005 M; the volume of S2O32- added = 10 mL. Therefore, moles of S2O32- added = (50 mL)*(1 L/1000 mL)*(0.005 mole/L) = 2.5*10-4 mole.

Now, the moles of S2O82- consumed have to be calculated as below.

moles of S2O82- consumed = (2.5*10-4 mole S2O32-)*(1 mole I2/2 mole S2O32-)*(1 mole S2O82-/1 mole I2) = 1.25*10-4 mole.

Therefore, the change in moles of S2O82- during the reaction = 1.25*10-4 M.

The rate of the reaction is

rate = -[S2O82-]/t = moles of S2O82- consumed/(L of reaction solution).(time) = (1.25*10-4 mole)/[(50 mL)*(1 L/1000 mL).(97 sec)] = (1.25*10-4 mole)*1000/[(50 L).(97 sec)] = 2.577*10-5 mole/L.sec 2.58*10-5 M/sec (ans).

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