Question

Does Q for the formation of 1 mol of ${\mathbf{NH}}_{\mathbf{3}}$from ${\mathbf{H}}_{\mathbf{2}}\mathbf{}{\mathbf{N}}_{\mathbf{2}}$differs from Q for the formation of ${\mathbf{NH}}_{\mathbf{3}}$from${\mathbf{H}}_{\mathbf{2}}$ and 1 mol of ${\mathbf{N}}_{\mathbf{2}}$? Explain and give the relationship between the two Q’s

Short answer

The answer is,

Yes, the reaction quotient in the first case shows half the stoichiometry of the reaction quotient in the second case.

Step-by-step solution

Reaction quotient

The Reaction quotient is represented using the letter Q. The reaction quotient is used for the relative amount of reactants and products in the reaction. The reaction quotient can be written using a balanced chemical equation.

Explanation

The chemical equation for the formation of 1 mol of${\mathrm{NH}}_3$ from ${\mathrm{H}}_2$and is given below.

$\frac{3}{2}{\mathrm{H}}_2+\frac{1}{2}{\mathrm{N}}_2\rightleftharpoons {\mathrm{NH}}_3$

The reaction quotient for the above reaction can be written as shown below.

role="math" localid="1657965009742" $\mathrm{Q}=\frac{\left[{\mathrm{NH}}_3\right]}{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

The chemical equation for the formation of ${\mathrm{NH}}_3$from ${\mathrm{H}}_2$and 1 mol of ${\mathrm{N}}_2$is given below.

$3{\mathrm{H}}_2+{\mathrm{N}}_2\rightleftharpoons 2{\mathrm{NH}}_3$

The reaction quotient for the above reaction can be written as shown below.

$\mathrm{Q}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$

The reaction quotient in both cases is different. The reaction quotient in the first case shows half the stoichiometry of the reaction quotient in the second case.