Chapter 13, Problem 13.8QP

Consider the reaction

${\text{N}}_2(g)+3{\text{H}}_2(g)\stackrel{}{\to }2{\text{NH}}_3(g)$

Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0.074 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 ${\text{NH}}_{\text{3}}$ being formed in the given reaction has to be determined from the given information.

Concept introduction:

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Rate of the reaction is the change in the concentration of reactant or a product with time.

For a general reaction,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A,B, C and D over time period $\Delta t$

Answer to Problem 13.8QP

The rate at which ${\text{NH}}_3$ being formed in the given reaction is $\text{0}\text{.049 M/s}$

Explanation of Solution

The given reaction is,

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

We have to calculate the rate of ammonia in the above reaction is as follows

$\text{rate = }-\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}=-\frac{1}{3}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}=\frac{1}{2}\frac{\text{Δ}\left[{\text{NH}}_3\right]}{\text{Δt}}$

$\frac{\text{Δ}\left[{\text{NH}}_3\right]}{\text{Δt}}=-\frac{2}{3}\left(-0.074\text{M/s}\right)\text{= 0}\text{.049 M/s}$

Interpretation Introduction

Interpretation:

The rate at which molecular nitrogen reacts in the given reaction has to be determined from the given information.

Concept introduction:

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Rate of the reaction is the change in the concentration of reactant or a product with time.

For a general reaction,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A,B, C and D over time period $\Delta t$

Answer to Problem 13.8QP

The rate at which molecular nitrogen reacts in the given reaction is $-0.025\text{M/s}$

Explanation of Solution

The given reaction is,

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

The given reaction

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

We have to calculate rate of nitrogen concentration in the above reaction is as follows

$\text{rate = }-\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}=-\frac{1}{3}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}=\frac{1}{2}\frac{\text{Δ}\left[{\text{NH}}_3\right]}{\text{Δt}}$

$\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}=-\frac{1}{3}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}=\frac{1}{3}\left(-0.074\text{M/s}\right)=-0.025\text{M/s}$