Consider the reaction scheme:

$$\ce{S + E ->[k_1] C1} \qquad \ce{C1 ->[k_2] E + P} \qquad \ce{S + C1 <=>[k_3][k_4] C2}$$

where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme-substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have used the law of mass reaction to convert this reaction into a system of differential equations:

\begin{align} \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2\\ \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\\ \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2\\ \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2\\ \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \end{align}

I am asked to find a conservation equation and use it to simplify this system however I haven't been taught as to what a conservation equation is. Can anyone push me in the correct direction? (and check the above differential equations are correct).


2 Answers 2


The reaction scheme is:

$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$

Let's define a bit of nomenclature.

  • S = compound S in chemical reaction equation
  • s = $[S]_x$, the concentration of S at t=x.
  • $S^*$ = ds/dt

The whole set of differential equations has 5 equations with 5 unknowns.

\begin{align} \\ S^* = \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2 \tag{4} \\ E^* = \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\tag{5} \\ C^*_1 = \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2 \tag{6} \\ C^*_2 = \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2 \tag{7}\\ P^* = \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \tag{8}\end{align}

What is a conservation equation

In mathematical terms a conservation equation is some linear combination of the time derivatives which add up to zero on the right hand side.

The gist is that the various conservation equations allow some simplification in the various interactions in the system.

Rather than just search blindly, there are a couple of relationships that can be exploited to find time derivatives which add up to zero on the right hand side. First is a sort of mass balance. The initial concentration of the reactants has to be spread as the various intermediate species and the final product. A second line of attack is to exploit and equilibrium conditions such as equation (3).

First conservation equation from "mass balance"

The first conservation equation is found noting that at t=0 then all is [S], a constant. Noting that it takes 2 $\ce{S}$ to make one $\ce{C_2}$, at some later time t=x, thus:

$$[\ce{S}]_{t=0} = [\ce{S}]_{t=x} + [\ce{C_1}]_{t=x} + 2[\ce{C_2}]_{t=x} + [\ce{P}]_{t=x}\tag{9} $$.

Now for some intermediate time we can take derivative of the mass balance equation with respect to time. All the dt's get cumbersome so lets use just $S^*$ from now on to mean ds/dt. From the mass equation then since the initial concentration is a constant we get

$0 = S^* + C^*_1 + 2C^*_2 + P^*\tag{10} $

Most likely solving for either $S^*$, or $P^*$ would give an interesting result.

A second conservation equation from mass balance

The second conservation equation comes from noting the enzyme participation in the reaction.

$$[\ce{E}]_{t=0} = [\ce{E}]_{t=x} + [\ce{C_1}]_{t=x} + [\ce{C_2}]_{t=x}\tag{11}$$.

Taking the derivative we get:

$ 0 = E^* + C^*_1 + C^*_2 \tag{12}$

which can be rearranged to

$E^* = - C^*_1 - C^*_2\tag{13}$

Substituting equation 6 for $C^*_1$ and equation 7 for $C^*_2$ this gives us equation 5 again which isn't helpful.

A third conservation equation from the equilibrium condition

From the third reaction we can derive the equation.

$ \frac{k_3}{k_4}= \frac{[S]_x[C_1]_x}{[C_4]_x} = \frac{sc_1}{c_4}\tag{14}$

Taking the derivative: $ 0 = \frac{c_1}{c_4}S^* + \frac{s}{c_4}C^*_1 - \frac{sc_1}{c_4^2}C^*_4\tag{15}$


$ S^* = -\frac{s}{c_1}C^*_1 + \frac{s}{c_4}C^*_4\tag{16}$

Infinite time as a boundary condition

Obviously at $t=\infty$, all the S has been converted to all P which is a constant. This factoid isn't very useful though in predicting how long that an "effective infinite time would be. Rather it is just a boundary condition.

$$[\ce{P}]_{t=\infty} = [\ce{S}]_{t=0}\tag{17}$$

  • $\begingroup$ This is not correct because the time derivative of s +c1+c2+p is not equal to zero. $\endgroup$ Nov 2, 2015 at 17:00
  • $\begingroup$ fixed equation because it takes 2S to make a C2. $\endgroup$
    – MaxW
    Nov 2, 2015 at 21:04
  • $\begingroup$ As I said before, and as the differential equations indicate, the quantity that stays constant is $(s+e+2c_1+3c_2 +p)$, so $\Delta (s+e+2c_1+3c_2 +p)=0$ $\endgroup$ Nov 2, 2015 at 22:51
  • $\begingroup$ Ok, I believe you on what the "conservation equation" is. (Question 1) What I have is the "mass balance" equation right? (Question 2) Can't I take the derivative with respect to time on that equation?!? $\endgroup$
    – MaxW
    Nov 2, 2015 at 22:57
  • 1
    $\begingroup$ I think you mean conservation equation. It's any linear combination of species concentrations that stays constant as the reactions progress. In this reaction scheme, there are two linearly independent combinations that remain constant. So, for this sequence, there is not just one THE conservation equation. There are two. Some of the coefficients in the linear combination are allowed to be zero, so the conserved quantities do not have to involve all 5 species. This all leads me to wonder whether there are other conserved combinations. I doubt it, but maybe. $\endgroup$ Nov 3, 2015 at 4:56

You need to find a linear combination of the time derivatives such that they add up to zero on the right hand side. This will tell you the corresponding linear combination of concentrations that is conserved. For this problem $s+e+2c_1+3c_2+p$ is conserved.

  • $\begingroup$ ah... I'm guessing that the "conservation equation" gives dP/dt? $\endgroup$
    – MaxW
    Nov 2, 2015 at 17:53
  • $\begingroup$ No. The conservation equation says that the quantity $(s+e+2c_1+3c_2+p)$ is conserved during the chemical reaction because its time derivative $\frac{d(s+e+2c_1+3c_2+p)}{dt}$ is equal to zero. $\endgroup$ Nov 2, 2015 at 18:37
  • $\begingroup$ Ok, I think I get it now. I gave equation for "mass balance" which yields one set of linear equations, the "conversation equation" yields another, and using them both you can get a simplified expression for dP/dt. // I need to correct my mass balance equation too. there are two S in C2. $\endgroup$
    – MaxW
    Nov 2, 2015 at 20:43
  • $\begingroup$ @ChesterMiller How does this equation simplify the system? $\endgroup$ Nov 4, 2015 at 20:42
  • $\begingroup$ The way that this simplifies the system is explained in the comments to MaxW's answer. MaxW came up with a second conservation equation that also applies to this problem. Basically, the conservation equations allow you to directly solve for s and e in terms of the concentrations of the other 3 species, and thereby allows you to reduce the number of differential equations from 5 to 3. $\endgroup$ Nov 4, 2015 at 22:23

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