Suppose, a reaction is like,

$$\ce{aA + bB <=> cC + dD}$$

and that A, B, C, D all are gas.

Now it is known that:

$$K_c = \dfrac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now if the forward reaction is second order and the reverse reaction is in first-order somehow then, how will this change.

I read some articles online but it was not actually clear that if all the reactants or products are in the same state like gas or aqueous then how $K_c$ or $K_p$ (for gas) will change depending on the rate of reaction or if they don't depend on that totally.

It will be helpful if one example is provided.

  • 1
    $\begingroup$ I'm not sure I understand but would not $K_c=k_f/k_r=[C]/([A][B])$ be an example, where the concentrations are the equilibrium values and $k_f, k_r$ are rate forward and reverse constants. (Normally for gas phase species we would use $K-p$ and use partial pressures instead of concentration.) $\endgroup$
    – porphyrin
    Feb 15, 2021 at 9:11
  • $\begingroup$ I'll point out that the equilbrium expression is inverted. For $K_c$ it should be reactants over products. $\endgroup$
    – MaxW
    Feb 17, 2021 at 0:44

1 Answer 1


I'll try and answer your question, but a "full" answer would take a book.

Given the reaction:

$$\ce{aA + bB <=> cC + dD}$$

Then assuming an elementary reaction in the gaseous state the concentration equilibrium constant always has products over reactants and will be:

$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

The above equation relies on two very specific assumptions:

  • The reaction is an elementary reaction which often isn't true and the actual coefficients must be determined experimentally since the reaction occurs in steps.
  • That concentrations can be used instead of activities.

Formally using activities explains part of your question immediately. For instance if A, B and C are gases and D is a liquid or solid, then the activity of D would be unity by definition.

Now breaking down the equilibrium expression into two rate equations we have:

  • $r_\mathrm{f}$ - Forward reaction rate
  • $r_\mathrm{r}$ - Reverse reaction rate
  • $\ce{a^*, b^*, c^*}$ and $\ce{d^*}$ = experimentally determined coefficients which may or may not be equal to the stoichiometric coefficients
  • $k_\mathrm{f}$ and $k_\mathrm{r}$ are constants for the forward and reverse reactions respectively.

Now using concentrations instead of activities:

\begin{align} r_\mathrm{f} &= k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}}\\ r_\mathrm{r} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}} \end{align}

and at equilibrium by definition:

\begin{align} r_\mathrm{f} &= r_\mathrm{r}\\ k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}} \end{align}


$$K_c = \frac{k_\mathrm{f}}{k_\mathrm{r}}= \frac{\ce{[C]^{c^*}[D]^{d^*}}}{\ce{[A]^{a^*}[B]^{b^*}}}$$

To determine the coefficients the experimentalist can manipulate the experiment to simply the kinetic expression. For example let $\ce{[A] \gg [B]}$ then $\ce{[A]^{a^*}}$ is essentially a constant.

Does this answer your question?

  • 1
    $\begingroup$ Should you not have $K_c=k_f/k_r=..$ in the last eqn instead of $K_c=r_r/r_f..$ ? $\endgroup$
    – porphyrin
    Feb 15, 2021 at 9:07
  • $\begingroup$ @porphyrin - You are correct. Thank you for pointing out the error. I've fixed my mistake. $\endgroup$
    – MaxW
    Feb 15, 2021 at 19:10
  • $\begingroup$ My question was what if the reverse reaction is the first-order reaction (suppose it depends on C). Then $K_c$ = $\frac{[C]^c}{[A]^a[B]^b}$ because as you wrote $r_f$=$r_r$ or not? $\endgroup$
    – user104796
    Feb 15, 2021 at 19:59
  • $\begingroup$ @user104796 - yes at equilibrium $\mathrm{r_f} = \mathrm{r_r}$. (1) for the reverse reaction to be first order it would have to depend on $\ce{[C]}$, not $\ce{[C]^c}$. (2) to eliminate $\ce{[D]^d}$ then make a reaction mixture of A, B and D where $\ce{[D] \gg [A] + [B] }$ so that $\ce{[D]^d}$ is constant and the reverse reaction would be pseudo-first order. $\endgroup$
    – MaxW
    Feb 15, 2021 at 20:22
  • $\begingroup$ I have got it. It helps me a lot. Thank you. $\endgroup$
    – user104796
    Feb 15, 2021 at 22:01

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