Chapter 18, Problem 56GQ

(a)

Interpretation Introduction

Interpretation:

The value of ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$, ${\text{ΔS}}^{\text{o}}\left(\text{system}\right)$ and ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for formation of methanol should be determined.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{system}\right){\text{=Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change for the reaction.

(b)

Interpretation Introduction

Interpretation:

It should be identified that the given reaction is product favoured at equilibrium or not at ${25}^{\text{o}}\text{C}$.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{system}\right){\text{=Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change for the reaction.