# Explanation for the derivative of kinetics equation

I'm working with a first rate law expression.

A + B -> C

Given: $$k_1[A][B]=k_{-1}[C]$$ which means $$[C]=\frac{k_1[A][B]}{k_-1}$$. The book says $$\frac{d[C]}{dt}=k_1[A][B]-k_{-1}[C]$$ which is where I'm lost on how to get to that step. I'm looking for an explanation into how this derivative is obtained. Thank you.

• It is not a mathematically derived derivative of a different equation, but rather a consequence of the mass balance which is also valid when the system is not at equilibrium. In other words, the forward reaction increases C, the reverse reaction decreases C. The concentration change at any moment depends on the two reaction rates and the concentrations of the species at that time. – MaxW Oct 15 '18 at 4:16
• So calculus techniques are not used (like quotient rule is what I was thinking), but it's just understood the mass of reactants has to equal mass of products? – Drake Oct 15 '18 at 4:18
• Sorry, I edited my answer while you were replying. Does the edited answer make sense? – MaxW Oct 15 '18 at 4:19
• Note that at equilibrium d[c] /dt = 0. So you want changes in C as system moves towards equilibrium. – MaxW Oct 15 '18 at 4:21

Your equation should really be written as $$\ce{A + B \rightleftharpoons C}$$ to indicate that both forwards and backwards reactions can occur as indicated by $$d[C]/dt = k_1[A][B]-k_{-1}[C]$$. This is normally the starting point. Only because of the reverse reaction $$\mathrm{C\to A+B}$$ is equilibrium is possible. At equilibrium the rate of change of C is zero so that $$k_1[A]_e[B]_e=k_{-1}[C]_e$$ and then the equilibrium constant is $$\displaystyle K_e=\frac{[C]_e}{[A]_e[B]_e}=\frac{k_1}{k_{-1}}$$. You can see how this relates to your second equation.

Often the $$e$$ subscripts are not added in the expressions which can lead to confusion. It is possible to work out the integrated rate expression for this scheme in terms of the amount reacted at any time and the amounts of A and B present initially.

Stealing bits from porphyrin I'll pose...

Your equation should really be written as: $$\ce{A + B <=>[k_1][k_{-1}] C}$$ to indicate that both forwards and backwards reactions can occur. Now we're interested in two different situations. First what is happening before equilibrium ids reached, and secondly what happens at equilibrium.

1. Before equilibrium

Let's use $$\ce{[A]_t}$$, $$\ce{[B]_t}$$, and $$\ce{[C]_t}$$ to denote the instantaneous concentrations of species A, B, and C at some time $$t$$. Now at time $$t$$, C is increasing in concentration due to the forward reaction and that rate is given by $$\ce{k_1[A]_t[B]_t}$$. At time $$t$$ C is also decreasing in concentration due to the backward reaction and that rate is given by $$\ce{k_{-1}[C]_t}$$. Hence the net change in C at time $$t$$ is given by $$\dfrac{d\ce{[C]}}{d\mathrm{t}} = \ce{k_1[A]_t[B]_t - k_{-1}[C]_t}$$

1. At equilibrium

The rate of change of C at equilibrium is zero so that $$k_1[A]_e[B]_e=k_{-1}[C]_e$$ and then the equilibrium constant is $$\displaystyle K_e=\frac{[C]_e}{[A]_e[B]_e}=\frac{k_1}{k_{-1}}$$.