2
$\begingroup$

In a reversible reaction $\ce{2NO2 <=>[$k_1$][$k_2$] N2O4}$, the rate of disappearance of $\ce{NO2}$ is equal to:

  1. $2\frac{k_1}{k_2}\ce{[NO2]}^2$
  2. $2k_1\ce{[NO2]}^2-2k_2\ce{[N2O4]}$
  3. $2k_1\ce{[NO2]}^2-k_2\ce{[N2O4]}$
  4. $(2k_1-k_2)\ce{[NO2]}$

The answer, they say, is (2). I just don't understand how they got it. I find it difficult to solve these questions. Is the rate of disappearance the derivative of the concentration of the reactant divided by its coefficient in the reaction, or is it simply the derivative?

If someone could help me with the solution, it would be great. And please, don't assume I'm just picking up a random question from a book and asking it for fun without actually trying to do it. I have worked at it and I don't understand what to do.

$\endgroup$
7
$\begingroup$

In general, if you have a system of elementary reactions, the rate of appearance of a species $\ce{A}$ will be

$$\cfrac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = \sum\limits_i \nu_{\ce{A},i} r_i$$

where

  • $i$ is each reaction in the system

  • $\nu_{\ce{A},i}$ is the stoichiometric coefficient of species $\ce{A}$ in reaction $i$ (positive for products, negative for reagents)

  • $r_i$ is the rate for reaction $i$, which in turn will be calculated as a product of concentrations for all reagents $j$ times the kinetic coefficient $k_i$:

$$r_i = k_i \prod\limits_{j} [j]^{\nu_{j,i}}$$

In other words, there's a positive contribution to the rate of appearance for each reaction in which $\ce{A}$ is produced, and a negative contribution to the rate of appearance for each reaction in which $\ce{A}$ is consumed, and these contributions are equal to the rate of that reaction times the stoichiometric coefficient.

The rate of disappearance will simply be minus the rate of appearance, so the signs of the contributions will be the opposite.

In your example, we have two elementary reactions:

$$\ce{2NO ->[$k_1$] N2O4} \tag{1}$$

$$\ce{N2O4 ->[$k_2$] 2NO} \tag{2}$$

So, the rate of appearance of $\ce{N2O4}$ would be

$$\cfrac{\mathrm{d}\ce{[N2O4]}}{\mathrm{d}t} = r_1 - r_2 $$

Similarly, the rate of appearance of $\ce{NO}$ would be

$$\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = - 2 r_1 + 2 r_2$$

and the rate of disappearance of $\ce{NO}$ would be minus its rate of appearance:

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 r_1 - 2 r_2$$

Since the rates for both reactions would be

$$r_1 = k_1 \ce{[NO]}^2$$

$$r_2 = k_2 \ce{[N2O4]}$$

the rate of disappearance for $\ce{NO}$ will be

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k_1 \ce{[NO]}^2 - 2 k_2 \ce{[N2O4]}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.