Question

Question: The alkali metals react with chlorine to give salts:

$$\begin{gathered} \mathbf{2}\mathbf{\text{Li}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{LiCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{2}\mathbf{\text{Na}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{NaCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{2}\mathbf{\text{K}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{KCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{2}\mathbf{\text{Rb}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{RbCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \end{gathered}$$

And $\mathbf{2}\mathbf{\text{Cs}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{CsCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$

Using the data in Appendix D, compute$∆\mathrm{S}°$ of each reaction and identify a periodic trend, if any.

Short answer

The change of entropy for given reaction are shown below:
a)$∆\mathrm{S}{°}_{\mathrm{a}}=-162.54{\mathrm{JK}}^{-1}$

b)$∆\mathrm{S}{°}_{\mathrm{b}}=-181.12{\mathrm{JK}}^{-1}$

c)$∆\mathrm{S}{°}_{\mathrm{c}}=-186.14{\mathrm{JK}}^{-1}$

d)$∆\mathrm{S}{°}_{\mathrm{d}}=-184.72{\mathrm{JK}}^{-1}$

e)$∆\mathrm{S}{°}_{\mathrm{e}}=-191.08{\mathrm{JK}}^{-1}$

Step-by-step solution

Given data

The value of standard entropies is given below:
$$\begin{gathered} S°\left(\mathrm{Li}\right(\mathrm{s}\left)\right)=29.12{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ S°({\mathrm{Cl}}_2(\mathrm{g}\left)\right)=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°(\mathrm{LiCl}(\mathrm{s}\left)\right)=59.33{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

Concept of the standard entropy

The entropy content of one mole of pure substance in a standard condition of pressure and any temperature of interest is known as the standard molar entropy in chemistry. The standard temperature and pressure are frequently chosen.

(a) Calculation of change of entropy for 2Li(s)+Cl2(g)→2LiCl(s)

The change in entropy is calculated with the help of the formula:
$∆\mathrm{S}°=\underset{}{\sum \mathrm{S}{°}_{\mathrm{products}}}-\underset{}{\sum \mathrm{S}{°}_{\mathrm{reactants}}}$ .......(i)

Where, $\underset{}{\sum \mathrm{S}{°}_{\mathrm{productios}}}$entropy of the product and $\sum \mathrm{S}°{}_{reac\tan ts}$is the entropy of the recant.

Put the value of given data in equation (i).
a)
$$\begin{gathered} ∆\mathrm{S}{°}_{\mathrm{a}}=2\mathrm{S}°(LiCl(s\left)\right)-2\mathrm{S}°(Li(s\left)\right)-\mathrm{S}°\left({\mathrm{Cl}}_2\right(\mathrm{g}\left)\right) \\ =2\mathrm{mol}\mathrm{x}59.33{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-2\mathrm{mol}\mathrm{x}29.12{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-1\mathrm{mol}\mathrm{x}222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ =118.66{\mathrm{JK}}^{-1}-58.24{\mathrm{JK}}^{-1}-222.96{\mathrm{JK}}^{-1} \\ =-162.54{\mathrm{JK}}^{-1} \end{gathered}$$

(b)Calculation of change of entropy for 2Na(s)+Cl2(g)→2NaCl(s)

It is given that,
$$\begin{gathered} \mathrm{S}°\left(\mathrm{NaCl}\right(\mathrm{s}\left)\right)=72.13{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°\left(C{l}_2\left(g\right)\right)=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°\left(NaCl\right(s\left)\right)=72.13{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

It can be calculated with the help of equation (i).

Calculation of change of entropy is shown below:
$$\begin{gathered} ∆\mathrm{S}{°}_b=2\mathrm{S}°\left(\mathrm{NaCI}\right(\mathrm{s}\left)\right)-2\mathrm{S}°\left(\mathrm{Na}\right(\mathrm{s}\left)\right)-\mathrm{S}°({\mathrm{Cl}}_2(\mathrm{g}\left)\right) \\ =2\mathrm{mol}\mathrm{x}72.59{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-2\mathrm{mol}\mathrm{x}51.21{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-1\mathrm{mol}\mathrm{x}222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ =144.26{\mathrm{JK}}^{-1}-104.42{\mathrm{JK}}^{-1}-222.96{\mathrm{JK}}^{-1} \\ =-181.14{\mathrm{JK}}^{-1} \end{gathered}$$

(c) Calculation of change of entropy for 2K(s)+Cl2(s)→2KCl(s)

It is given that,

$$\begin{gathered} \mathrm{S}°\left(\text{K}\right(s\left)\right)=64.18{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{Cl}}_2(\mathrm{g}\left)\right)=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°\left(\mathrm{KCI}\right(\mathrm{s}\left)\right)=82.59{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$
It can be calculated with the help of equation (i).

Calculation of change of entropy is shown below:
$$\begin{gathered} ∆\mathrm{S}{°}_{\mathrm{c}}=2\mathrm{S}°\left(\mathrm{KCI}\right(\mathrm{s}\left)\right)-2\mathrm{S}°\left(\mathrm{K}\right(\mathrm{s}\left)\right)-\mathrm{S}°({\mathrm{Cl}}_2(\mathrm{g}\left)\right) \\ =2\mathrm{mol}\mathrm{x}82.59{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-2\mathrm{mol}\mathrm{x}64.18{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-1\mathrm{mol}\mathrm{x}222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ =165.18{\mathrm{JK}}^{-1}-128.36{\mathrm{JK}}^{-1}-222.96{\mathrm{JK}}^{-1} \\ =-186.14{\mathrm{JK}}^{-1} \end{gathered}$$

(c) Calculation of change of entropy for 2Rb(s)+Cl2(s)→2RbCl(s)

It is given that,
$$\begin{gathered} \mathrm{S}°\left(\mathrm{Rb}\right(s\left)\right)=76.78{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{Cl}}_2(\mathrm{g}\left)\right)=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°\left(\mathrm{RbcI}\right(\mathrm{s}\left)\right)=95.90{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

It can be calculated with the help of equation (i).

Calculation of change of entropy is shown below:
$$\begin{gathered} ∆\mathrm{S}{°}_{\mathrm{d}}=2\mathrm{S}°\left(\mathrm{RbCl}\right(s\left)\right)-2\mathrm{S}°\left(\mathrm{Rb}\right(s\left)\right)-\mathrm{S}°\left({\mathrm{Cl}}_2\right(\mathrm{g}\left)\right) \\ =2\mathrm{mol}\mathrm{x}95.90{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-2\mathrm{mol}\mathrm{x}76.78{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-1\mathrm{mol}\mathrm{x}222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ =191.80{\mathrm{JK}}^{-1}-153.56{\mathrm{JK}}^{-1}-222.96{\mathrm{JK}}^{-1} \\ =-184.72{\mathrm{JK}}^{-1} \end{gathered}$$

(e) Calculation of change of entropy for 2Cs(s)+Cl2(s)→2CsCl(s)

It is given that,
$$\begin{gathered} \mathrm{S}°\left(\mathrm{Cs}\right(s\left)\right)=85.23{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{Cl}}_2(\mathrm{g}\left)\right)=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°\left(\mathrm{CsCI}\right(\mathrm{s}\left)\right)=101.17{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

It can be calculated with the help of equation (i).

Calculation of change of entropy is shown below:
$$\begin{gathered} ∆\mathrm{S}{°}_{\mathrm{e}}=2\mathrm{S}°\left(\mathrm{CsCl}\right(s\left)\right)-2\mathrm{S}°\left(\mathrm{Cs}\right(s\left)\right)-\mathrm{S}°\left({\mathrm{Cl}}_2\right(\mathrm{g}\left)\right) \\ =2\mathrm{mol}\mathrm{x}101.17{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}-2\mathrm{mol}\mathrm{x}85.23{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ =202.34{\mathrm{JK}}^{-1}-170.56{\mathrm{JK}}^{-1}-222.96{\mathrm{JK}}^{-1} \\ =-191.08{\mathrm{JK}}^{-1} \end{gathered}$$