Chapter 23, Problem 23.39P

Interpretation Introduction

Interpretation:

Value of $\Delta G_{rxn}^o$, $\Delta H_{rxn}^o$ and $\Delta S_{rxn}^o$ has to be determined for the reaction ${\text{H}}_2\left(g\right){\text{ + CO}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }{\text{H}}_{2}O\left(g\right)\text{+ CO}\left(g\right)$. Also, the factor that drives the reaction and the direction in which the reaction is spontaneous under standard conditions has to be identified.

Concept Introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\text{ΔG}$.

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

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

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.

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:

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

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

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

Answer to Problem 23.39P

Value of $\Delta G_{rxn}^o$ is $28.6\text{ kJ/mol}$.

Value of $\Delta H_{rxn}^o$ is $41.2\text{ kJ/mol}$.

Value of $\Delta S_{rxn}^o$ is $\text{0}\text{.0}42\text{kJ}/\text{K}\cdot \text{mol}$.

It is an entropy favored reaction

Explanation of Solution

The given reaction is shown below,

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

• Calculate the value of $\Delta G_{rxn}^o$:

Standard free energy of formation values is given below,

  $\begin{array}{c}\text{Δ}G°\left[{\text{H}}_2\left(g\right)\right]\text{=}0\text{ kJ/mol}\\ \\ \text{Δ}G°\left[{\text{CO}}_{\text{2}}\left(g\right)\right]\text{=}-\text{394}\text{.4 kJ/mol}\\ \\ \text{Δ}G°\left[{\text{H}}_{2}O\left(g\right)\right]\text{=}-\text{228}\text{.6 kJ/mol}\\ \\ \text{Δ}G°\left[\text{CO}\left(g\right)\right]\text{=}-\text{137}\text{.2 kJ/mol}\end{array}$

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

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

Substitute the values as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{= }\left[\left(1\right)\left(-\text{228}\text{.6 kJ/mol}\right)+\left(1\right)\left(-\text{137}\text{.2 kJ/mol}\right)\right]\\ -\left[\left(1\right)\left(0\text{ kJ/mol}\right)+\left(1\right)\left(-\text{394}\text{.4 kJ/mol}\right)\right]\\ \\ =28.6\text{ kJ/mol}\end{array}$

Therefore, the value of $\Delta G_{rxn}^o$ is $28.6\text{ kJ/mol}$.

• Calculate the value of $\Delta H_{rxn}^o$:

Standard enthalpy of formation values is given below,

  $\begin{array}{c}\text{Δ}H°\left[{\text{H}}_2\left(g\right)\right]\text{=}0\text{ kJ/mol}\\ \\ \text{Δ}H°\left[{\text{CO}}_{\text{2}}\left(g\right)\right]\text{=}-\text{393}\text{.5 kJ/mol}\\ \\ \text{Δ}H°\left[{\text{H}}_{2}O\left(g\right)\right]\text{=}-\text{241}\text{.8 kJ/mol}\\ \\ \text{Δ}H°\left[\text{CO}\left(g\right)\right]\text{=}-\text{110}\text{.5 kJ/mol}\end{array}$

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

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

Substitute the values as follows,

  $\begin{array}{c}\Delta H_{rxn}^o\text{= }\left[\left(1\right)\left(-\text{241}\text{.8 kJ/mol}\right)+\left(1\right)\left(-\text{110}\text{.5 kJ/mol}\right)\right]\\ -\left[\left(1\right)\left(-\text{393}\text{.5 kJ/mol}\right)+\left(1\right)\left(\text{0 kJ/mol}\right)\right]\\ \\ =41.2\text{ kJ/mol}\end{array}$

Therefore, the value of $\Delta H_{rxn}^o$ is $41.2\text{ kJ/mol}$.

• Calculate the value of $\Delta S_{rxn}^o$:

Standard entropy of formation values is given below,

  $\begin{array}{c}\text{Δ}S°\left[{\text{H}}_2\left(g\right)\right]\text{=}\text{130}\text{.7 }\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[{\text{CO}}_{\text{2}}\left(g\right)\right]\text{=}\text{213}\text{.8 }\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[{\text{H}}_{2}O\left(g\right)\right]\text{=}\text{188}\text{.8 }\text{J}/\text{K}\cdot \text{mol}\\ \\ \text{Δ}S°\left[\text{CO}\left(g\right)\right]\text{=}\text{197}\text{.7 }\text{J}/\text{K}\cdot \text{mol}\end{array}$

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,

  $\begin{array}{c}\Delta S_{rxn}^o\text{= }\left[\left(1\right)\left(\text{188}\text{.8 }\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\left(\text{197}\text{.7 }\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ -\left[\left(1\right)\left(\text{130}\text{.7 }\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\left(\text{213}\text{.8 }\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ \\ =42\text{J}/\text{K}\cdot \text{mol}\\ \\ =\text{0}\text{.0}42\text{kJ}/\text{K}\cdot \text{mol}\end{array}$

Therefore, value of $\Delta S_{rxn}^o$ is $\text{0}\text{.0}42\text{kJ}/\text{K}\cdot \text{mol}$.

Here, $\Delta S_{rxn}^o>0$. Hence, it is an entropy favored reaction.

Also, $\Delta G_{rxn}>0$ so the reaction is spontanous in backward reaction and the direction of reaction from right to left.