Chapter 14, Problem 8QP

Consider the reaction:

${\text{N}}_{\text{2}}\text{(}g\text{)+3}{\text{H}}_{\text{2}}\text{(}g\text{)}\to \text{2N}{\text{H}}_{\text{3}}\text{(}g\text{)}$

Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of $\text{0}\text{.082 }M/s$. (a) At what rate is ammonia being formed? (b) At what rate is molecular nitrogen reacting?

Interpretation Introduction

Interpretation:

The rate at which ammonia reacts and at which nitrogen reacts with given rate of reaction of molecular hydrogen is to be calculated.

Concept introduction:

The branch of chemistry thatdeals with the relative rates of the chemical reactions and the factors affecting the rates of reactions is known as chemical kinetics.

The rate of reaction is the measurement of concentration change of reactants within a period of time.

For the reaction: $\text{A}+3\text{B}\to 2\text{C}$.

If three moles of B are reacted then its rate expression will be divided by 3and if two moles of C are formed then its rate expression will be divided by 2

The rate expression of reactant has negative sign because they are consumed in reaction.

The rate expression of product has positive sign because they are formed in reaction.

Rate is always a positive quantity, the concentration of reactant decreases, a minus sign is required in rate expression to make it positive.

The rate expression for the reaction is given by:

$-\frac{\Delta [A]}{\Delta t}=-\frac{1}{3}\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}$

Answer to Problem 8QP

Solution: (a) $0.055M{s}^{-1}$,

(b) $-0.027M{s}^{-1}$.

Explanation of Solution

a) Rate of formation of ammonia

The given reaction is as follows:

$N_{2(g)}+3H_{2(g)}\to 2N{H}_{3(g)}$

Molecular hydrogen reacts at the rate of $0.082M{s}^{-1}$.

The rate expression for the reaction is given as follows:

$-\frac{\Delta [N_2]}{\Delta t}=-\frac{1}{3}\frac{\Delta [H_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [N{H}_3]}{\Delta t}$

The rate at which ammonia is formed is calculated by the expression as follows:

$\begin{array}{c}-\frac{1}{3}\frac{\Delta [H_2]}{\Delta t}=+\frac{1}{2}\frac{\Delta \left[N{H}_3\right]}{\Delta t}\\ \frac{\Delta \left[N{H}_3\right]}{\Delta t}=-\frac{2}{3}\frac{\Delta [H_2]}{\Delta t}\end{array}$

Substitute the value of rate at which molecular hydrogen reacts in the above equation.

$\begin{array}{c}\frac{\Delta [H_2]}{\Delta t}=-\frac{2}{3}\left(-0.082M{s}^{-1}\right)\\ =0.055M{s}^{-1}\end{array}$

The rate at which ammonia is formed is $0.055M{s}^{-1}$.

Given information:

The given reaction is as follows:

$N_{2(g)}+3H_{2(g)}\to 2N{H}_{3(g)}$

Molecular hydrogen reacts at the rate of $0.082M{s}^{-1}$.

b) The rate at molecular nitrogen reacts.

The rate expression for the reaction is given as follows:

$-\frac{\Delta [N_2]}{\Delta t}=-\frac{1}{3}\frac{\Delta [H_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [N{H}_3]}{\Delta t}$

The rate at whichnitrogen reacts is calculated by the expression as follows:

$\frac{\Delta [N_2]}{\Delta t}=\frac{1}{3}\frac{\Delta [H_2]}{\Delta t}$

Substitute the value of rate at which molecular hydrogen reacts in the above equation,

$\begin{array}{c}\frac{\Delta [N_2]}{\Delta t}=\frac{1}{3}\left(-0.082M{s}^{-1}\right)\\ =-0.027M{s}^{-1}\end{array}$

The molecular nitrogen reacts at the rate of $-0.027M{s}^{-1}$.