Chapter 23, Problem 23.21P

Interpretation Introduction

Interpretation:

Value of $\Delta S_{rxn}^o$ and $\Delta G_{rxn}^o$ has to be calculated for the reaction ${\text{C}}_{\text{2}}{\text{H}}_4\left(g\right)+\text{ 3}{\text{O}}_2\left(g\right)\underset{}{\rightleftharpoons }2{\text{ CO}}_2\left(g\right){\text{ + 2 H}}_{2}O\left(g\right)$.  The direction in which the reaction is spontaneous has to be indicated.

Concept Introduction:

Entropy$\left(S\right)$: it is used to describe the disorder.  It is the amount of arrangements possible in a system at a particular state.

Value of $\Delta S_{rxn}^o$ can be calculated by the equation:

  $\Delta S_{rxn}^o\text{= }S°\left[\text{products}\right]-S°\left[\text{reactants}\right]$

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\Delta G$. Value of $\Delta G_{rxn}^o$ is determined as follows,

  $\Delta G_{rxn}^o=\Delta H_{rxn}^o-T\Delta S_{rxn}^o$

The formula of ${\text{ΔG}}_{\text{rxn}}$ in terms of thermodynamic reaction quotient (Q),

    ${\text{ΔG}}_{rxn}={\text{ΔG}}_{rxn}^o+\text{RT}\text{ln Q}$

If $\Delta G_{rxn}<0$ then the reaction is spontanous in forward reaction and so the reaction will favor products.

If $\Delta G_{rxn}>0$ then the reaction is spontanous in backward reaction and so the reaction will favor reactants.

If $\Delta G_{rxn}=0$ then the reaction is at equilibrium.

Answer to Problem 23.21P

Value of $\Delta S_{rxn}^o$ is $-29.7\text{J}/\text{K}\cdot \text{mol}$.

Value of $\Delta G_{rxn}^o$ is $-1314\text{ kJ/mo}l$.

The reaction is spontaneous in forward direction.

Explanation of Solution

Given information is shown below,

  $\begin{array}{l}T\text{ = }25°C=\left(25+273\right)=298K\\ \\ \Delta H_{rxn}^o\text{ =}-\text{1323}\text{.0 kJ/mol}\end{array}$

The given reaction is shown below,

  ${\text{C}}_{\text{2}}{\text{H}}_4\left(g\right)+\text{ 3}{\text{O}}_2\left(g\right)\underset{}{\rightleftharpoons }2{\text{ CO}}_2\left(g\right){\text{ + 2 H}}_{2}O\left(g\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}\text{Δ}S°\left[{\text{C}}_{\text{2}}{\text{H}}_4\left(g\right)\right]\text{=}\text{219}\text{.3}\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[{\text{O}}_2\left(g\right)\right]\text{=}\text{205}\text{.2}\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[{\text{CO}}_2\left(g\right)\right]\text{=}\text{213}\text{.8}\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[{\text{H}}_{2}O\left(g\right)\right]\text{=}\text{188}\text{.8}\text{J}/\text{K}\cdot \text{mol}\end{array}$

Value of $\Delta S_{rxn}^o$ can be calculated by the equation:

  $\Delta S_{rxn}^o\text{= }S°\left[\text{products}\right]-S°\left[\text{reactants}\right]$

Substitute the values as follows,

$\begin{array}{c}\Delta S_{rxn}^o\text{= }\left[\left(2\right)\left(\text{213}\text{.8}\text{J}/\text{K}\cdot \text{mol}\right)+\left(2\right)\left(\text{188}\text{.8}\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ -\left[\left(1\right)\left(\text{219}\text{.3}\text{J}/\text{K}\cdot \text{mol}\right)+\left(3\right)\left(\text{205}\text{.2}\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ \\ =-29.7\text{J}/\text{K}\cdot \text{mol}\\ \\ =-\text{0}\text{.0297}\text{kJ}/\text{K}\cdot \text{mol}\end{array}$

Therefore, value of $\Delta S_{rxn}^o$ is $-29.7\text{J}/\text{K}\cdot \text{mol}$.

Value of $\Delta G_{rxn}^o$ is determined as follows,

  $\begin{array}{c}\Delta G_{rxn}^o=\Delta H_{rxn}^o-T\Delta S_{rxn}^o\\ \\ =\left(-\text{1323}\text{.0 kJ/mo}l\right)-\left(298K\times -\text{0}\text{.0297}\text{kJ}/\text{K}\cdot \text{mol}\right)\\ \\ =-1314\text{ kJ/mo}l\end{array}$

Therefore, value of $\Delta G_{rxn}^o$ is $-1314\text{ kJ/mo}l$.

Thermodynamic reaction quotient is determined as follows,

  $\begin{array}{c}Q\text{ = }\frac{{\left({\text{P}}_{{\text{CO}}_2}\right)}^2{\left({\text{P}}_{{\text{H}}_{2}O}\right)}^2}{\left({\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_4}\right){\left({\text{P}}_{{\text{O}}_2}\right)}^3}\\ \\ =\frac{{\left(\text{20}\text{.0}\right)}^2{\left(\text{0}\text{.010}\right)}^2}{\left(\text{0}\text{.010}\right){\left(\text{0}\text{.020}\right)}^3}\\ \\ =\text{5}\text{.0}\times {\text{10}}^5\end{array}$

Value of ${\text{ΔG}}_{\text{rxn}}$ is determined as follows,

  $\begin{array}{c}{\text{ΔG}}_{rxn}={\text{ΔG}}_{rxn}^o+\text{RT}\text{ln Q}\\ \\ \text{=}\left(-1314\text{ kJ/mo}l\right)+\left[\left(\text{8}\text{.314}\times {\text{10}}^{-3}\text{ kJ}{\text{.K}}^{-1}.mo{l}^{-1}\right)\left(298\text{ K}\right)\text{ln }\left(\text{5}\text{.0}\times {\text{10}}^5\right)\right]\\ \\ =-\text{1281 kJ/mol}\end{array}$

Therefore, value of ${\text{ΔG}}_{\text{rxn}}$ is $-\text{1281 kJ/mol}$.

Here, $\Delta G_{rxn}<0$ and it indicates that the reaction is spontaneous in forward direction. Direction of reaction is from left to right.