Use the ${\mathrm{pK}}_{\mathrm{a}}$table in Appendix B to determine in which direction the

Equilibrium is favored.

(c)localid="1650537382286" $\dot{\mathbf{C}}{\mathbf{H}}_{\mathbf{3}}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\rightleftarrows }{\mathbf{CH}}_{\mathbf{4}}\mathbf{+}\dot{\mathbf{C}}{\mathbf{H}}_{\mathbf{2}}{\mathbf{NO}}_{\mathbf{2}}$

Nitromethane is an organic compound with a molecular formula $\left({\mathrm{CH}}_3{\mathrm{NO}}_2\right)$. The boiling point of nitromethane is , and in chemical industries, it is used for extraction, as a cleaning solvent, etc.

$\left({\mathrm{CH}}_3{\mathrm{NO}}_2\right)$

${\mathrm{pK}}_{\mathrm{a}}$ Value = 10.21

Hydrocarbons are of two types: saturated and unsaturated. Saturated hydrocarbons have single bonds, and unsaturated hydrocarbons have double and triple bonds. Methane is a saturated hydrocarbon having one carbon atom, and four hydrogen atoms are attached to it to complete the valency of the carbon atom.

${\mathrm{CH}}_3$

${\mathrm{pK}}_{\mathrm{a}}$ Value = 60

As the ${\mathrm{pK}}_{\mathrm{a}}$value of methane is higher than the ${\mathrm{pK}}_{\mathrm{a}}$value of nitromethane, the acidic strength of the acid and base pair of the product side is less than the acidic strength of the reactant side. We know that equilibrium shifts in that direction in which acidic strength is less. Thus, the equilibrium will shift on the product side of the reaction.

$\dot{\mathrm{C}}{\mathrm{H}}_3+{\mathrm{CH}}_3{\mathrm{NO}}_2\rightleftarrows {\mathrm{CH}}_4+\dot{\mathrm{C}}{\mathrm{H}}_2{\mathrm{NO}}_2$