Chapter 19, Problem 19.68QE

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction has to be written and the amount of carbon needed to produce $\text{22}{\text{.8 L CO}}_{\left(\text{g}\right)}\text{ at 1}\text{.00 atm}$ and $\text{298 K}$ has to be calculated.  $\text{ΔH,}\text{ΔS}\text{and}\text{ΔG}$ has to be calculated at $\text{298 K}$ and also the minimum temperature needed to make this reaction spontaneous at standard-state conditions $\text{ΔG}\text{<}\text{0}$ has to be estimated.

Explanation of Solution

The balanced equation for the given reaction is,

  ${\text{MnO}}_{\text{2}}\text{+}{\text{2C}}_{\text{(s)}}\stackrel{}{\to }{\text{Mn}}_{\text{(s)}}\text{+}{\text{2CO}}_{\text{(g)}}$

The amount of carbon needed to produce $\text{22}{\text{.8 L CO}}_{\left(\text{g}\right)}\text{ at 1}\text{.00 atm}$ and $\text{298 K}$ is calculated by using ideal gas equation.

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\end{array}$

Substitute the given values in above equation to get moles of $\text{22}\text{.8 L CO}$.

  $\text{=}\frac{\text{1×22}\text{.8}}{\text{0}\text{.082×298}\text{K}}\text{=}\text{0}\text{.932}\text{mol}$

From the above balanced equation, 1 mole carbon produced 1 mole $\text{CO}$ therefore, mole the amount of carbon needed to produce $\text{22}\text{.8 L CO}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.932}\text{mol×12}\text{.0}\text{g}\\ \text{=}\text{11}\text{.184}\text{g}\end{array}$

Hence, $\text{11}\text{.184}\text{g}$ carbon is needed to produce $\text{22}{\text{.8 L CO}}_{\left(\text{g}\right)}\text{ at 1}\text{.00 atm}$ and $\text{298 K}$.

The enthalpy change of the reaction is calculated by using Hess’s law,

  ${\text{ΔH}}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}$

Substitute the standard values in above equation to get enthalpy change of the reaction.

  $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}\\ \text{=2(-110}\text{.52)-(-}\text{520}\text{.03)}\\ \text{=(-221}\text{.04)-(-}\text{520}\text{.03)}\\ \text{=}\text{295}\text{.99}\text{kJ/mol}\end{array}$

Enthalpy change of the reaction is $\text{295}\text{.99}\text{kJ/mol}$.

Entropy change of reaction is,

$\begin{array}{l}\text{ΔS}\text{=}\frac{\text{ΔH}}{\text{T}}\\ \text{=}\frac{\text{295}\text{.99}\text{kJ/mol}}{\text{298}\text{K}}\text{=0}\text{.99}\text{kJ/mol}\cdot \text{K}\end{array}$

Gibbs free energy change of the reaction is,

  $\begin{array}{l}\text{ΔG}\text{=}\text{ΔH-TΔS}\\ \text{=}\text{295}\text{.99}\text{kJ/mol-298K}\text{(0}\text{.99}\text{kJ/mol}\text{×K)}\\ \text{=}\text{295}\text{.99}\text{kJ/mol-295}\text{.02}\text{kJ/mol}\\ \text{=}\text{0}\text{.97}\text{kJ/mol}\end{array}$

The minimum temperature needed to make this reaction spontaneous at standard-state conditions $\text{ΔG}\text{<}\text{0}$.

  $\begin{array}{l}\text{ΔG}\text{=}\text{ΔH-TΔS}\\ \text{=}\text{295}\text{.99}\text{kJ/mol-299K}\text{(0}\text{.99}\text{kJ/mol}\text{×K)}\\ \text{=}\text{295}\text{.99}\text{kJ/mol-296}\text{.01kJ/mol}\\ \text{=}-\text{0}\text{.02}\text{kJ/mol}\\ \text{=}-\text{20}\text{J/mol}\end{array}$

Hence, $\text{299K}$ is the minimum temperature that needed to make this reaction spontaneous at standard-state conditions $\text{ΔG}\text{<}\text{0}$.