Chapter 19.3, Problem 19.2WE

Consider the reaction

${\text{4NO}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to {\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(}g\text{)}$

At a particular time during the reaction, nitrogen dioxide is being consumed at the rate of 0.00130 M/s. (a) At what rate is molecular oxygen being consumed? (b) At what rate is dinitrogen pentoxide being produced?

Interpretation Introduction

Interpretation:

At what rate is molecular oxygen being consumed has to be determined.

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.

Explanation of Solution

To solve this problem first calculate the rate constant for the given reaction and then determine the rate from the obtained rate constant.

The given reaction is ${\text{4NO}}_{\text{2}}{\text{+O}}_{\text{2}}\to {\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}$

The rate expression for the above reaction is as follows

$\text{rate = -}\frac{1}{4}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=-\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}=\frac{1}{2}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}{\text{O}}_5\right]}{\text{Δt}}$

From the given data,

$\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=-0.00130M/s$

Where the negative sign indicates that the concentration of ${\text{NO}}_{\text{2}}$ is decreasing with time.

Therefore, the rate of the given reaction is

$\text{rate = -}\frac{1}{4}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=\text{-}\frac{1}{4}(-0.00130M/s)$

$\text{rate =}3.25\times {10}^{-4}M/s$

The rate for molecular oxygen is determined as follows

$3.25\times {10}^{-4}M/s=\frac{\text{Δ}\left[{\text{O}}_2\right]}{\text{Δt}}$

$\frac{\text{Δ}\left[{\text{O}}_2\right]}{\text{Δt}}=-3.25\times {10}^{-4}M/s$

Interpretation Introduction

Interpretation:

At what rate is dinitrogen pentoxide being produced has to be determined.

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.

Explanation of Solution

To solve this problem first calculate the rate constant for the given reaction and then determine the rate from the obtained rate constant.

The given reaction is ${\text{4NO}}_{\text{2}}{\text{+O}}_{\text{2}}\to {\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}$

The rate expression for the above reaction is as follows

$\text{rate = -}\frac{1}{4}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=-\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}=\frac{1}{2}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}{\text{O}}_5\right]}{\text{Δt}}$

From the given data,

$\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=-0.00130M/s$

Where the negative sign indicates that the concentration of ${\text{NO}}_{\text{2}}$ is decreasing with time.

Therefore, the rate of the given reaction is

$\text{rate = -}\frac{1}{4}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=\text{-}\frac{1}{4}(-0.00130M/s)$

$\text{rate =}3.25\times {10}^{-4}M/s$ (b)

The rate for dinitrogen pentoxide is determined as follows

$3.25\times {10}^{-4}M/s=\frac{1}{2}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}{\text{O}}_5\right]}{\text{Δt}}$

$2\left(3.25\times {10}^{-4}M/s\right)=\frac{\text{Δ}\left[{\text{N}}_{\text{2}}{\text{O}}_5\right]}{\text{Δt}}$

$\frac{\text{Δ}\left[{\text{N}}_{\text{2}}{\text{O}}_5\right]}{\text{Δt}}=6.50\times {10}^{-4}M/s$