Chapter 13, Problem 13.150QP

Nitrogen dioxide reacts with carbon monoxide by the overall equation

${\text{NO}}_2(g)+\text{CO}(g)\stackrel{}{\to }\text{NO}(g)+{\text{CO}}_2(g)$

At a particular temperature, the reaction is second order in NO₂ and zero order in CO. The rate constant is 0.515 L/(mol · s). How much heat energy evolves per second initially from 3.50 L of reaction mixture containing 0.0275 M NO₂? See Appendix C for data. Assume the enthalpy change is constant with temperature.

Interpretation Introduction

Interpretation:

The amount of heat energy liberated by from 3.50L of solution per second has to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

$\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

$\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

$\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

To determine the amount of heat energy liberated

Answer to Problem 13.150QP

Answer

The amount of heat liberated is found to be $\text{-0}\text{.308 kJ/s}$.

Explanation of Solution

Given,

Rate constant for decomposition = $\text{0}\text{.515}\text{L/}\left(\text{mol}\text{.s}\right)$

Volume and Molarity of Hydrogen peroxide = $\text{3}\text{.50}\text{L}\text{\&}\text{0}\text{.0275}\text{M}$

The first order rate law is used to calculate the initial rate of decomposition given as,

Rate = $\text{k}{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}$

Rate = $\left(\text{0}\text{.515}\text{L/}\left(\text{mol}\text{.s}\right)\right)\text{×}{\left(\text{0}\text{.0275}\text{M}{\text{NO}}_{\text{2}}\right)}^{\text{2}}$

Rate= $\text{3}{\text{.894×10}}^{\text{-4}}\text{M}{\text{NO}}_{\text{2}}\text{/s}$

The heat liberated per second in per mol of ${\text{NO}}_{\text{2}}$ is calculated by calculating the standard enthalpy of decomposition of one mol of ${\text{NO}}_{\text{2}}$

$\begin{array}{l}{\text{NO}}_{\text{2(g)}}\text{+}{\text{CO}}_{\left(\text{g}\right)}\to {\text{NO}}_{\left(\text{g}\right)}+{\text{CO}}_{\text{2(g)}}\\ \\ {\text{ΔH°}}_{\text{f}}\text{=}\text{33}\text{.10}\text{kJ/mol}-\text{110}\text{.5}\text{kJ/mol}\text{90}\text{.29}\text{kJ/mol}\text{-393}\text{.5}\text{kJ/mol}\end{array}$

The change in standard enthalpy is

$\begin{array}{l}\text{ΔH°=90}\text{.29-393}\text{.5-}\left(\text{33}\text{.10}\right)\text{-}\left(\text{-110}\text{.5}\right)\\ \\ \text{ΔH°=-225}\text{.81 }\text{kJ/mol}\end{array}$

The heat liberated per second is calculated as,

$\frac{\text{-225}\text{.81}\text{kJ}}{\text{mol}{\text{NO}}_{\text{2}}}\text{×}\frac{\text{3}{\text{.894×10}}^{\text{-3}}\text{mol}{\text{NO}}_{\text{2}}}{\text{L}\text{.s}}\text{×3}\text{.50}\text{L=-0}\text{.3077=-0}\text{.308}\text{kJ/s}$

Conclusion

The amount of heat energy liberated by from 3.50 L of solution per second was calculated using rate of first order rate law and the change in standard enthalpy.  The amount of heat liberated per second was found to be $\text{-0}\text{.308 kJ/s}$.