Question

Using an $ICE$ Table. How do we calculate Ice for the following reaction? At $373{\text{ }}K$, $0.1{\text{ }}mols\;$ of ${N_2}{O_4}$ is heated in a one-liter flask and at equilibrium the amount of nitrogen dioxide is $0.12{\text{ }}mols$.

Answer 1

An ICE (Initial, Change, Equilibrium) table is straightforward matrix formalism that is used to rearrange the figuring in reversible balance responses (e.g., powerless acids and frail bases or complex particle arrangement). ICE tables naturally set up and sort out the factors and constants required while computing the unknown.

Complete step by step answer:
First off, we (obviously) expect ideal gases (these are the two gases at room temperature).
At that point, we review the meaning of the concentration equilibrium constant for a two-segment balance:
${K_c} = \dfrac{{{{\left[ B \right]}^{{\nu _B}}}}}{{{{\left[ A \right]}^{{\nu _A}}}}}$,
for harmony .
Despite the fact that we are seeing two gases, we are given the mols and the all-out vessel's volume, so we can at present get starting and balance fixations of course.
In any case, since the vessel won't change size, and ideal gases are accepted to fill the vessel totally and equally, let us essentially work in mols for the ICE table, and gap by $1{\text{ }}L$ later.
Ordinarily, we would compose this:

${\text{I}}\;\:\;\:\;\:\;\:0.1\;\:\;\:\;\:\;\:\;\:\;\:0$
${\text{C}}\;\:\;\:\;\:\;\: - x\;\:\;\: + 2x$
${\text{E}}\;\:\;\:0.1 - x\;\:\;\:\;\:\;\:2x$
taking note of that the utilization of $x{\text{ }}mols$ of ${N_2}{O_4}$yields $2x{\text{ }}mols$ of $N{O_2}\left( g \right)$, and not simply $x$.
Be that as it may, we understand what $x$ is. Since at balance, ${n_{N{O_2}}} = {\text{0}}{\text{.12 mols}}$, we have that $2x = {\text{0}}{\text{.12 mol}}s$, or $x = {\text{0}}{\text{.06 mols}}$. Thusly, we would already be able to figure ${K_c}$:
${K_c} = \dfrac{{{{\left[ {N{O_2}} \right]}^2}}}{{\left[ {{N_2}{O_4}} \right]}}$
$= \dfrac{{{{\left( {2x} \right)}^2}\;\:{\text{mol}}{{\text{s}}^{\not{2}}}{\text{/}}\not{{\text{1}}}{{\text{L}}^{\not{2}}}}}{{\left( {0.1 - x} \right)\not{{{\text{mols}}}}{\text{/}}\not{{{\text{1 L}}}}}}$
$= \dfrac{{4{x^2}}}{{0.1 - x}}M$
$= \dfrac{{4{{\left( {0.06} \right)}^2}}}{{0.1 - 0.06}}M$
Be that as it may, ${K_c}$ is traditionally revealed without units, so:
${K_c} = 0.36$

Note:
ICE tables are made out of the groupings of particles in arrangement in various phases of a response, and are generally used to compute the ${K_c}$, or harmony steady articulation, of a response (in certain occasions, ${K_c}$ might be given, and at least one of the focuses in the table will be the obscure to be settled for).