Naked Science Forum
On the Lighter Side => Science Experiments => Topic started by: scientizscht on 04/08/2019 23:01:24

Hello
I have a graph that shows the concentration of a product increasing as a reaction takes place.
Let's say that graph is a straight link y=ax starting from zero at time = 0 and goes to 1M when time = 1 hour.
How can I calculate from that the amount of that product in the solution of 1L?
I was tempted to calculate the area under the line but that would give units of Mxh which does not make sense.
Any idea?

In general, reaction rates are not a straight line.
In a closed container, as the reactants become more dilute, the reaction rates decline.
A negativeexponential rate is more realistic.

In general, reaction rates are not a straight line.
In a closed container, as the reactants become more dilute, the reaction rates decline.
A negativeexponential rate is more realistic.
I know but the whole point of the question is different.
As I said, let's assume a straight line.
How can the amount of the product be calculated from that graph?

What does 1M mean in this?
goes to 1M when time = 1 hour.

What does 1M mean in this?
goes to 1M when time = 1 hour.
Molarity
(are you surely a chemist?)

I have a graph that shows the concentration of a product increasing as a reaction takes place.
So that well be moles/liter (concentration) on the y axis, and hours on the xaxis, and the concentration at any time will be the y value for that time.

What does 1M mean in this?
goes to 1M when time = 1 hour.
Molarity
(are you surely a chemist?)
I'm a chemist, so it's obvious that I know what it means.
So it's clear that I'm not asking you so that I can find out.
Now, let's try again
What does the M mean, (or if you like, what does "molarity" mean)?

I have a graph that shows the concentration of a product increasing as a reaction takes place.
Let's say that graph is a straight link y=ax starting from zero at time = 0 and goes to 1M when time = 1 hour.
How can I calculate from that the amount of that product in the solution of 1L?
I was tempted to calculate the area under the line but that would give units of Mxh which does not make sense.
Any idea?