[soupastupid]

im looking at
A --> B --> C
where k1 is A to B and k2 is B to C. (rate constants)

i'm having trouble understanding the rate law

d[A] / dt = -k1*[A]

why is k negative??
or can i just move the negative?

-d[A] / dt = k1*[A]  
to show that concentration of [A] is decreasing but the rate has to be positive


my next question is
for d[ B]/dt = k1[A] - k2[ B]

why does this equation have k1[A]??

my final question is
if k1 is bigger than k2

does [ B] accumulate faster than [C] is being produced?

thanks guys
-andrew

[Schrödinger]

d[A] / dt = -k1*[A]

why is k negative??
or can i just move the negative?

-d[A] / dt = k1*[A] 
to show that concentration of [A] is decreasing but the rate has to be positive

Yes you can move the negative sign. As a matter of fact, you can see the significance of the equation only in the latter form. As you rightly said, it is to show that [A] is decreasing since it is a reactant, and the rate has to be positive

my next question is
for d[ B]/dt = k1[A] - k2[ B]

why does this equation have k1[A]??

B is involved in 2 reactions. It is being formed from A and it also decomposes to C. So, there is an increase in its concentration thanks to A and decrease in its concentration as C is produced. So, [A] is bound to affect .

my final question is
if k1 is bigger than k2

does B accumulate faster than [C] is being produced?

Yes. The difference k₁[A]-k₂B decides the sign of dB/dt, thereby indicating whether there is net production of B or net consumption.

[rabolisk]

my next question is
for d[ B]/dt = k1[A] - k2[ B]

why does this equation have k1[A]??

In a reaction where A B , for every mole of A that is consumed, a mole of B must be produced. Hence if [A] decreases by some amount x, [ B] increases by some amount x. Taking the derivative, -d[A]/dt = d[ B]/dt. The negative sign shows that when [A] is decreasing, [ B] is increasing. Since -d[A]/dt = k1[A], d[ B]/dt = k1[A].

Now you have a second reaction, B C. With only this reaction, d[ B]/dt = k2[ B]. Since B participates in both, you add the two together.

my final question is
if k1 is bigger than k2

does [ B] accumulate faster than [C] is being produced?

thanks guys
-andrew

It depends on the concentration of A and B. When you have some A, but no B, then B has to accumulate faster than C is produced, because you have no B to produce C, regardless of the values of k1 and k2. When you have no A, but some B, then B isn't accumulating at all, and C is of course being produced, at least to some extent. When k1 >>> k2, most of A will be converted to B, before C can be produced. On the other hand, when k1 <<< k2, you will see very low levels of [ B], because it will be converted to C as soon as it appears. You can see this mathematically.