Rate Constant - Help Needed

I did an experiment investigating how concentration affects the rate of reaction between sodium thiosulphate and hydrochloric acid.

I worked out the orders to give the following equation to work out the rate constant:

rate constant = rate / [ sodium thiosulphate] [ hydrochloric acid ]

(where rate = 1/time taken)

In the experiment I changed the concentration of sodium thiosulphate while keeping hydrochloric acid the same, and changed the concentration of hydrochloric acid while keeping the concentration of sodium thiosulphate the same.

The problem is this gave me two rate constants. So what would the rate constant 'k' be in the following equation?

rate = k [Sodium Thiosulphate] [ Hydrochloric Acid]

The values of both rate constants I got when rounded were 0.18. So does k = 0.18? I'm just confused because I worked out two rate constants (one for sodium thiosulphate and one for hydrochloric acid)

Thanks in advance

Reply 1

charco

Study Forum Helper

18

you cant work out a rate constant for the reaction without knowing the orders with respect to the reactants. The first parts of your experiment you should have worked out the orders then, using the data from any one experiment, you work out the rate constant

Reply 2

jonnyq

OP

charco

you cant work out a rate constant for the reaction without knowing the orders with respect to the reactants. The first parts of your experiment you should have worked out the orders then, using the data from any one experiment, you work out the rate constant

Yeah sorry I forgot to include that I did work out the orders.

rate constant = rate / [ sodium thiosulphate] [ hydrochloric acid ]

The [sodium thiosulphate] and [hydrochloric acid] are both to the power of one in the equation above, so both are first order with respect to each one.

Reply 3

Kyri

15

The way you described working out k seems correct to me. You say that you got two different values, but both round to 0.18. This is understandable as you worked this out experimentally and because of this, there is always going to be some sort of minor error involved in certain things such as timing or measurements during titrations. So I would say if what you have done is correct, which it seems to be, k does indeed equal 0.18.

Edit: Another thing to add, the reason for the slight difference in values of k might well have been that both reactions weren't done at the exact same temperature. Temperature affects the rate of reaction so any difference at all, even very small, would affect the rate constant. But I still think you should take it as 0.18, that's probably pretty much as accurate as you can be.

Reply 4

jonnyq

OP

Ok thanks for clearing that up for me. I appreciate the help.

One more thing, is there a formula I can use to calculate the half life. I know it can be worked out by looking at the graph, I just want to know if there's a mathematical formula. I think the formula relates to the orders.

Thanks again.

Reply 5

Kyri

15

Half-life can be calculated with the formula:

half-life = ln2 / k

(k = rate constant)

In case you don't know, ln2 is natural log 2, and is equal to approxmiately 0.69.

Kyri

Half-life can be calculated with the formula:

half-life = ln2 / k

(k = rate constant)

In case you don't know, ln2 is natural log 2, and is equal to approxmiately 0.69.

This can ONLY be used for a first order reaction.

Reply 7

Kyri

15

charco

This can ONLY be used for a first order reaction.

Yes, you are absolutely right. I completely overlooked that, sorry. If the overall order of the reaction is not 1 then the half-life is not constant so the equation does not work. So unless there is another formula I am not aware of, for reaction orders above 1 I think you have to use the graph to find half-life, which won't be the same value throughout the reaction.