[Boxxxed]

If I have

Rate = 5.69x10-5
Concentration at time 0 = 0.021 mol/L
Temp= 293 k

Would the value for concentration at time t be 0 because this is when the reaction is complete?

How do I find k constant?

This is a lab where we timed the reaction between Na2S2O3 and HCl

[rabolisk]

Concentration of what?

What is the reaction? How many reactants are there? Do you know the rate law for this reaction?
Also, a reaction may only proceed to equilibrium. Reactants could still be remaining, so the concentration may not be 0 even at infinite time.

[Boxxxed]

concentration of Na2S2O3


 Dropping 5 ml of 2 mol/L HCl makes hcl the limiting reagent.

0.0005 mol of Na2S2O3 reacts with 0.001 mol HCl.

Don't know the rate law because I don't know the reaction order. That is what I need to figure out to find the constant.

[rabolisk]

Write out the full reaction. I assume that you're manually timing your reaction and looking for change in color? This color change signifies some sort of chemical change. You need to give me a bit more information.

[Boxxxed]

Reaction is

S2O3+2H--->H2O+SO2+S

S is the precipitate that I was looking for. I would stop the timer when the solution became opaque.

HCl was mixed with a total of 5 different solutions made by serial dilution.

Solution A B C D and E, concentration of A starting at 27g/mol (Na2S2O3*5H2O) and lowered by a dilution factor of 1.5 for each solution.

so B=18g/mol
    C=12g/mol etc.

With lowering concentration the time it took for the reaction to take place increased.

So what I have is the reaction time and concentrations for those 5 solutions.

I also have information regarding solution E (0.021 mol/L) and the time for the reaction to take place with it heated at different temperatures.

I am not sure how to find the rate constant. I am currently in the process of graphing In[C] vs time and 1/[C] time and I have found that 1/[C] vs time to be linear ignoring the deviations due to experimental error. So is the reaction order 2?

[Boxxxed]

What would the rate law be?