Question

129. Given the thermodynamic data below, calculate $\mathbf{∆}\mathbf{S}$ and$\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$ for the following reaction at 25°C and 1 atm:
${\mathbf{XeF}}_{\mathbf{6}}\left(g\right)\mathbf{\to }{\mathbf{XeF}}_{\mathbf{4}}\left(s\right)\mathbf{+}{\mathbf{F}}_{\mathbf{2}}\left(g\right)$

	$\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\left(\mathrm{kJ}/\mathrm{mol}\right)$	${\mathbf{S}}^{\mathbf{o}}\mathbf{}\left({\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)$
${\mathbf{XeF}}_{\mathbf{6}}\mathbf{}\left(g\right)$	-294	300
${\mathbf{XeF}}_{\mathbf{4}}\left(s\right)$	-251	146
${\mathbf{F}}_{\mathbf{2}}\left(g\right)$	0	203

Short answer

The$\mathbf{∆}\mathbf{S}$ and $\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$ for the given reaction at 25°C and 1 atm are$\mathbf{49}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$ and $\mathbf{-}\mathbf{144}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$.

Step-by-step solution

Step 1: Introduction to the Concept

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy, spontaneous change occurs.

The mathematical equation for the standard entropy value at room temperature is given as follows,

$\mathbf{∆}{\mathbf{S}}_{\mathbf{R}}^{\mathbf{o}}\mathbf{=}\mathbf{\sum }{\mathbf{n}}_{\mathbf{p}}\mathbf{∆}{\mathbf{S}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{\left(}\mathbf{Products}\mathbf{\right)}\mathbf{-}\mathbf{\sum }{\mathbf{n}}_{\mathbf{r}}\mathbf{∆}{\mathbf{S}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{\left(}\mathbf{Reactants}\mathbf{\right)}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

In the balanced chemical equation, n and p indicate the coefficients of reactants and products.

The following is the relationship betweenentropy change and enthalpy change for the surrounding:

$\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}\mathbf{=}\mathbf{-}\frac{\mathbf{∆}\mathbf{H}}{\mathbf{T}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(2\right)$

$∆{\mathrm{S}}_{\mathrm{surr}}=$Entropy change of the environment

$∆\mathrm{H}=$Enthalpy change

$\mathrm{T}=$Temperature

Step 2: Determination of ∆S at 25°C and 1 atm

The given reaction,

${\mathrm{XeF}}_6\left(\mathrm{g}\right)\to {\mathrm{XeF}}_4\left(\mathrm{s}\right)+{\mathrm{F}}_2\left(\mathrm{g}\right)$

By equation (1),

$$\begin{gathered} ∆{\mathrm{S}}^0=\left[1{\mathrm{S}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{XeF}}_4\left(\mathrm{s}\right)\right)+1{\mathrm{S}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{F}}_2\left(\mathrm{g}\right)\right)\right]-\left[1{\mathrm{S}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{XeF}}_6\left(\mathrm{g}\right)\right)\right] \\ ∆{\mathrm{S}}^0=\left[1\mathrm{mole}\times 146\mathrm{J}/\mathrm{molK}+1\mathrm{mole}\times 203\mathrm{J}/\mathrm{molK}\right]-\left[1\mathrm{mole}\times 300\mathrm{J}/\mathrm{molK}\right] \\ ∆{\mathrm{S}}^0=\left(146+203-300\right)\mathrm{J}/\mathrm{molK} \\ ∆{\mathrm{S}}^0=49\mathrm{J}/\mathrm{K} \end{gathered}$$

Step 3: Determination of ∆H at 25°C and 1 atm

By equation (1),

$$\begin{gathered} ∆{\mathrm{H}}^0=\left[1∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{XeF}}_4\left(\mathrm{s}\right)\right)+1∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{F}}_2\left(\mathrm{g}\right)\right)\right]-\left[1∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{XeF}}_6\left(\mathrm{g}\right)\right)\right] \\ ∆{\mathrm{H}}^0=\left[1\mathrm{mole}\times -251\mathrm{kJ}/\mathrm{mol}+1\mathrm{mole}\times 0\mathrm{kJ}/\mathrm{mol}\right]-\left[1\mathrm{mole}\times -294\mathrm{kJ}/\mathrm{mol}\right] \\ ∆{\mathrm{H}}^0=\left(-251+294\right)\mathrm{kJ}/\mathrm{mol} \\ ∆{\mathrm{H}}^0=43\mathrm{J}/\mathrm{K} \end{gathered}$$

Step 4: Determination of ∆Ssurr at 25°C and 1 atm

By equation (2),

$\begin{array}{rcl}\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}& \mathbf{=}& \mathbf{-}\frac{\mathbf{∆}\mathbf{H}}{\mathbf{T}}\\ & =& \frac{-\left(43000\mathrm{J}\right)}{298\mathrm{K}}\\ & =& -144\mathrm{J}/\mathrm{K}\\ & & \\ & & \\ & & \end{array}$

Therefore $\mathbf{∆}\mathbf{S}$is $49\mathrm{J}/\mathrm{K}$and $\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$is $-144\mathrm{J}/\mathrm{K}$.