Question

Give the units of ${\mathrm{K}}_{\mathrm{C}}$ for the formation of 1 mole of ${\mathrm{NH}}_3$ from its constituents.

Answer 1

• The reaction for the formation of 1 mole of Ammonia $\left({\mathrm{NH}}_3\right)$ is:

$\frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right)$

• For this reaction, ${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{NH}}_3\right]}{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \frac{3}{2}}}}$
• Putting the units of concentration in the above equation, units of ${\mathrm{K}}_{\mathrm{C}}\to \frac{\left(\mathrm{mol}/\mathrm{L}\right)}{{(\mathrm{mol}/\mathrm{L})}^{{\displaystyle \frac{1}{2}}}{(\mathrm{mol}/\mathrm{L})}^{{\displaystyle \frac{3}{2}}}}$ = $\frac{\left(\mathrm{mol}/\mathrm{L}\right)}{{\left(\mathrm{mol}/\mathrm{L}\right)}^2}$ = ${\left(\mathrm{mol}/\mathrm{L}\right)}^{-1}$ = $\mathrm{L}/\mathrm{mol}$ or $\mathrm{L}{\mathrm{mol}}^{-1}$
• Hence, the unit of ${\mathrm{K}}_{\mathrm{C}}$ for the formation of 1 mole of ${\mathrm{NH}}_3$ is Litre mol^-1.