Question

Question: Consider the following reaction:

${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{CO}\left(g\right)\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$

Amounts of H₂O, CO, H₂, and CO₂ are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive ¹⁴C, will ¹⁴C be found only in CO molecules for an indefinite period of time? Why or why not?

Short answer

Answer

${}^{14}\mathrm{C}$will be found in both carbon monoxide and carbon dioxide molecules because it is an equilibrium reaction.

Step-by-step solution

Equilibrium reaction

No, because it is an equilibrium reaction. We know that equilibrium is defined as a dynamic process.

Why carbon-14 is found for an indefinite period of time

The equation for the reaction between and is written below:

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+{}^{14}\mathrm{CO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{}^{14}{\mathrm{CO}}_2\left(\mathrm{g}\right)$

${}^{14}\mathrm{C}$will be found in both carbon monoxide and carbon dioxide molecules because it is an equilibrium reaction. In the reaction, some of ${}^{14}\mathrm{C}\mathrm{O}\left(\mathrm{g}\right)$is converted into ${}^{14}{\mathrm{CO}}_2\left(\mathrm{g}\right)$to re-establish the equilibrium.