# Rate law quick question??

Jul 25, 2017

Well, the rate, ${r}_{2} \left(t\right) = - \frac{1}{2} \frac{\Delta \left[E\right]}{\Delta t}$ (negative for reactants!) would not change, as long as the stoichiometry of the reaction did not change.

And since it does not, that does not change if reaction 2 were a non-fast step. You may be able to write ${r}_{1}$ in terms of ${r}_{2}$, if you knew those numerically, but if you do not, then you should note that $\frac{\Delta \left[D\right]}{\Delta t}$ is not necessarily the same between reactions $1$ and $2$.

The rate law, however, does change.

(As a sidenote, probably not the best example if you want to find a rate law!)

GETTING THE RATE LAW IF THE SECOND STEP IS FAST

Well, if the first step is the only slow step, it should give rise to a rate law dependent on mostly that first step, treating it as an elementary reaction:

$r \left(t\right) = k \left[A\right] {\left[B\right]}^{3}$

For this process, the overall reaction is apparently:

$\text{A" + 2"E" -> 2"C" + "F}$

with rates:

$r \left(t\right) = - \frac{1}{1} \frac{\Delta \left[A\right]}{\Delta t} = - \frac{1}{2} \frac{\Delta \left[E\right]}{\Delta t} = \frac{1}{2} \frac{\Delta \left[C\right]}{\Delta t} = \frac{1}{1} \frac{\Delta \left[F\right]}{\Delta t}$

But $B$ is a catalyst, not a reactant... So we would have to next eliminate $B$ in the rate law we have temporarily jotted down.

To do this, we would have used something called the steady-state approximation (SSA) on step 1, paired with the fast equilibrium approximation (FEA) on step 2.

• The SSA states that the step forming an intermediate is so slow that the step after it (if it is fast) consumes it immediately, and its change in concentration is effectively zero.
• The FEA states that the equilibrium is established almost right away, so that the equilibrium constant $K$ can be written.

If the second step is not fast, then we could not make the SSA. In that case, the true rate law would be a jumbled mess, with potentially fractional orders on $A$ and $E$, and a non-obvious observed rate constant.

The reason why we could have written $r \left(t\right) = k \left[A\right] {\left[B\right]}^{3}$ with a fast step 2 is because it was fast; we assume that step 2 is so fast, that it has practically no weight on the rate law, i.e. that the order with respect to reactant $E$ is effectively zero.

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TREATING THE FIRST STEP USING SSA

The SSA allows us to write:

$\frac{d \left[D\right]}{\mathrm{dt}} = {k}_{1} \left[A\right] {\left[B\right]}^{3} - {k}_{- 1} {\left[C\right]}^{2} \left[D\right] - {k}_{2} {\left[E\right]}^{2} \left[D\right] + {k}_{- 2} \left[F\right] {\left[B\right]}^{3} \approx 0$ $\text{ } \boldsymbol{\left(1\right)}$

detailing the contribution of each reaction step and direction to the overall change in concentration of $D$ over time. A negative subscript indicates the reverse reaction for that step.

TREATING THE SECOND STEP USING FEA

The FEA allows us to write:

$\frac{{r}_{2}}{{r}_{- 2}} = \frac{{k}_{2} {\left[E\right]}^{2} \left[D\right]}{{k}_{- 2} \left[F\right] {\left[B\right]}^{3}} = 1$ $\text{ } \boldsymbol{\left(2\right)}$

The equilibrium constant would be given by ${K}_{2} = \frac{\left[F\right] {\left[B\right]}^{3}}{{\left[E\right]}^{2} \left[D\right]}$, so at equilibrium, ${r}_{2} = {r}_{- 2}$, and:

$1 = {k}_{2} / \left({k}_{- 2}\right) \cdot \frac{1}{K} _ 2$

$\implies {K}_{2} = {k}_{2} / \left({k}_{- 2}\right)$ $\text{ } \boldsymbol{\left(3\right)}$

FINDING THE OVERALL RATE LAW?

Rearranging $\left(1\right)$:

${k}_{1} \left[A\right] {\left[B\right]}^{3} + {k}_{- 2} \left[F\right] {\left[B\right]}^{3} = {k}_{2} {\left[E\right]}^{2} \left[D\right] + {k}_{- 1} {\left[C\right]}^{2} \left[D\right]$

$\left[D\right] = \frac{{k}_{1} \left[A\right] {\left[B\right]}^{3} + {k}_{- 2} \left[F\right] {\left[B\right]}^{3}}{{k}_{2} {\left[E\right]}^{2} + {k}_{- 1} {\left[C\right]}^{2}}$

However, $B$ is a catalyst. So, we would have needed to find an expression for $B$, or already know its final concentration.

(And this process would be done until every intermediate or catalyst has been expressed as reactants. It is presumed you know what the concentrations of your products and catalysts are in an experiment.)