Which of the following is the correct expression for the rate of the following reaction? Note: not all species are gaseous. CuO(s) + H2S (g) <--> Cu(s) + H2O (g)

A. [Cu] x [H2O] / ([CuO] x [H2S]) B. [H2O] / [H2S] C. [CuO] x [

Aug 19, 2017

It would only include the non-solids and non-pure-liquids, i.e. aqueous species and gases. $\left(B\right)$.

(By the way, your reaction doesn't make physical sense. Your copper product should be copper(II) sulfide...)

Well, I think there is a typo in the question... While it's true that the equilibrium constant for the reaction

$a A + b B \stackrel{\text{ "k_1" }}{r i g h t \le f t h a r p \infty n s} c C + \mathrm{dD}$
${\text{ "" "" "" }}^{{k}_{- 1}}$

is derived by equating the rate laws of the forward and reverse reactions:

${r}_{1} \left(t\right) = {k}_{1} {\left[A\right]}^{a} {\left[B\right]}^{b}$ " "" "" "bb("(Forward reaction)")
${r}_{- 1} \left(t\right) = {k}_{- 1} {\left[C\right]}^{c} {\left[D\right]}^{d}$ " "" "bb("(Reverse reaction)")

${K}_{e q} = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}} = {k}_{1} / \left({k}_{- 1}\right)$,

... the given multiple choice answers are each equilibrium constant expressions, also called mass action expressions.

These describe a steady state (in which ${r}_{1} \left(t\right) = {r}_{- 1} \left(t\right)$), i.e. it's not going to appear to be going in any particular direction unless something disturbs it.

Equilibrium constants are written to include only aqueous and gaseous species. Solids and pure liquids are considered to have activities of $1$, and are omitted. So, for the reaction

$\text{CuO"(s) + "H"_2"S"(g) rightleftharpoons "CuS"(s) + "H"_2"O} \left(g\right)$,

which should be inspected and compared to the one given in the question (how did sulfur enter and not get produced again?), the solid copper(II) oxide and solid copper(II) sulfide (not elemental copper) are omitted, and we only write:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "K_(eq) = (["H"_2"O"])/(["H"_2"S"])" }}{|}}}}$

Or, since these are gases, if we had the partial pressures, we could have written:

$\underline{{K}_{P} = \left({P}_{\text{H"_2"O"))/(P_("H"_2"S}}\right)}$