Can anyone help me to make an empirical equation for this dissolution kinetics? Where Y is time, X is % of remaining material. X0 is an initial quantity. X1-3 are the starting points after time=0.1. T1-3 represent to different temperatures. Please let me know if you have any concern. Thank you in advance.
These are three parallel lines of the form x = m(Y - 0.1) + c, where c= X1-3 for T1-3. From the graphs one has to find the value of slope m.
Dear Prof. Mittal
Thank you so much. I have got an equation to explain those three lines but the problem is that how can I make this equation associate to X0?
X0 is associated with the slope and values X1-3 for temperature T1 - 3. Do experiments and relate X0 with X1 for T1 etc.
Hi Wirat,
your curves are lines but in semi-log plots, so that means your functions to fit is an exponential (and more precisely something close to 10^(aX+b) ) . So let's say y=f(X) is one of your curve before plotting it in the semi-log axes then:
Y=log10(y)=(aX+b) are your curve.
So you just need to find a and b as a classique linear regression. If I take the blue curve for example, and assume Xk is the value for which Y=k, then we have:
log10(aXk+b)=k i.e. aXk+b=10^k.
Just take two values of Xk and you can compute a and b. It should work fine.
Hi Prof. Sio-Hoi leng
Actually, the slope of each line is not parallelly aligned. They are divergent. However, what I have got; X = (mx+b)logY+(mx+b)--first (mx+b) is for slope and the rest for the intercept. But I am working on how can I start to to make an equation from X0? Could you help me more, thank you so much.
Wirat, I am not assuming any parallelism of your lines. I just consider one of them (e.g. blue one). You are estimating the slope a and the constant b by taking 2 arbitrary points from the same line:
Before the semi log plot, if y=10^(aX+b), then the semi log is plotting Y(X)=log10(y(X)).
Then if 1=log10(10^(aX1+b)) and 2=log10(10^(aX2+b)), you have:
a=(2-1)/(X2-X1), thus b=1-aX1 = 2-aX2
So for any value X and X', your parameters are:
a=(Y(X)-Y(X'))/(X-X') and b=Y(X)-aX.
So you got your answer.
Prof. Sio-Hoi Leng
Thank you so much..I will try my best to do that again.
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