Solving Law of Mass Action Problem with X', Y', and Z

[Carla1985]

I am working through some questions in a book and have got stuck on one. The question is:

"The equations below come from applying the law of mass action to two reactions.

$$X'=aXY \\
Y'=bYZ+cZ \\
Z'=dYZ+eZ$$

Find the two reactions and determine how the coefficients a,b,c,d and e are related, if at all. Assume a,b,c,d,e are nonzero but can be positive or negative."

I have got the first reaction as $X+Y\overset{a}\rightarrow Y$ with being negative. For the second reaction I initially tried $Y+Z\rightleftarrows Z$ which would work for $Y'$ but doesn't for $Z'$.

Could someone please point me in the direction of how I can work out what this second reaction needs to be.

Thanks
Carla

 

[I like Serena]

I am working through some questions in a book and have got stuck on one. The question is:

"The equations below come from applying the law of mass action to two reactions.

$$X'=aXY \\
Y'=bYZ+cZ \\
Z'=dYZ+eZ$$

Find the two reactions and determine how the coefficients a,b,c,d and e are related, if at all. Assume a,b,c,d,e are nonzero but can be positive or negative."

I have got the first reaction as $X+Y\overset{a}\rightarrow Y$ with being negative. For the second reaction I initially tried $Y+Z\rightleftarrows Z$ which would work for $Y'$ but doesn't for $Z'$.

Could someone please point me in the direction of how I can work out what this second reaction needs to be.

Thanks
Carla

Hey Carla! ;)

It seems to me that it works for Z' as well.
Why do you think that it doesn't? (Wondering)

 

[Carla1985]

Hey Carla! ;)

It seems to me that it works for Z' as well.
Why do you think that it doesn't? (Wondering)

My supervisor explained it to me that if I have a reaction, $Y+aZ\rightarrow bZ$ then the differential equation for $Z$ is proportional to $(b-a)YZ$, so if $a=b$ the difference is $0$ so the coefficient is $0$, i.e. there is no change in the concentration of $Z$.

That was my understanding. Have I misunderstood somewhere along the way?

 

[I like Serena]

My supervisor explained it to me that if I have a reaction, $Y+aZ\rightarrow bZ$ then the differential equation for $Z$ is proportional to $(b-a)YZ$, so if $a=b$ the difference is $0$ so the coefficient is $0$, i.e. there is no change in the concentration of $Z$.

That was my understanding. Have I misunderstood somewhere along the way?

I see a couple of differences.
The reaction is in 2 directions instead of 1.
And it is not given that a=b, so we should assume they are different. (Thinking)

 

[Carla1985]

Sorry, I'm a little confused. If I have the reaction I originally thought $Y+Z\leftrightarrows Z$ isn't $Z'$ just $0$ as there is a $Z$ on either side of the reaction so the amount of $Z$ doesn't change with the reaction?

 

[I like Serena]

Sorry, I'm a little confused. If I have the reaction I originally thought $Y+Z\leftrightarrows Z$ isn't $Z'$ just $0$ as there is a $Z$ on either side of the reaction so the amount of $Z$ doesn't change with the reaction?

We don't know.
We can only assume that Z decreases proportionally to YZ, and simultaneously increases due to the right hand side.
And then Z decreases proportionally to Z, and simultaneously increases due to the left hand side.
To model this, I think we should say that $Z'=dYZ+eZ$, which fits the equations.

 

[I like Serena]

I have got the first reaction as $X+Y\overset{a}\rightarrow Y$ with being negative.

This is a bit odd actually.
We are assuming that the concentration of Y on the left side decreases with the same amount that it increases on the right side, so that effectively there is no change in Y.
But if that is the case, where did X go? (Wondering)
It seems to contradict conservation of mass.