Question

Question: All of the halogens reaction directly with to give binary compounds. The reactions are

$$\begin{gathered} {\mathbf{\text{F}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{HF}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ {\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{HCl}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ {\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{HBr}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{\text{I}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{HI}} \end{gathered}$$

Using the data in Appendix D, compute of each reaction and identify a periodic trend, if any.

Short answer

The change of entropy of a given reaction is written as:
A: $∆\mathrm{S}°=14.1\mathrm{J}/\mathrm{Kmol}$
B: $∆\mathrm{S}°=20.07\mathrm{J}/\mathrm{Kmol}$
C:$∆S°=21.26\mathrm{J}/\mathrm{Kmol}$
D: $∆\mathrm{S}°=21.81\mathrm{J}/\mathrm{Kmol}$

Step-by-step solution

Given data

The value of standard entropies is given below:
$$\begin{gathered} \mathrm{S}°({\mathrm{H}}_2,\mathrm{g})=130.57{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{F}}_2,\mathrm{g})=202.67{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°(\mathrm{HF},\mathrm{g})=173.67{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

Concept of the standard entropy

The randomness, chaos, or freedom of movement of particles is measured by entropy. Entropy is represented by the capital letter S . However, you won't find simple "entropy" calculations because the notion isn't useful unless it's placed into a form that can be utilized to conduct comparisons and compute a change in entropy or S. The entropy of one mole of a substance under standard state circumstances is expressed as standard molar entropy. The symbol for standard molar entropy is $\mathbf{S}\mathbf{°}$ , and the units are generally joules per mole Kelvin.

(a) Calculation of change of entropy for F2(g)+H2(g)→2HF(g)

The change in entropy is calculated with the help of the formula:

$∆\mathrm{S}°=\sum \mathrm{S}{°}_{\mathrm{products}}-\sum \mathrm{S}{°}_{\mathrm{reactants}}$ …..(i)

Where,entropy of the product and is the entropy of the recant.

Put the value of given data in equation (i).

$$\begin{gathered} ∆\mathrm{S}°=2\times \mathrm{S}°(\mathrm{HF},\mathrm{g})-\mathrm{S}°\left({\mathrm{F}}_{2,}\mathrm{g}\right)-\mathrm{S}°\left({\mathrm{H}}_{2,}\mathrm{g}\right) \\ ∆\mathrm{S}°=2\times 173.67-202.67-130.57 \\ ∆\mathrm{S}°=14.1\mathrm{J}\hspace{0.17em}/\mathrm{Kmol} \end{gathered}$$

(b) Calculation of change of entropy for Cl2(g)+H2(g)→2HCl(g)

It is given that,
$$\begin{gathered} \mathrm{S}°\left({\mathrm{H}}_{2,}\mathrm{g}\right)=130.57{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{Cl}}_2\mathrm{g})=222.96{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°(\mathrm{HCI},\mathrm{g})=186.80{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

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

Calculation of change of entropy is shown below:

$$\begin{gathered} ∆\mathrm{S}°=2\times \mathrm{S}°(\mathrm{HCI},\mathrm{g})-\mathrm{S}°({\mathrm{Cl}}_{2,}\mathrm{g})-\mathrm{S}°({\mathrm{H}}_2,\mathrm{g}) \\ ∆\mathrm{S}°=2\times 186.80-222.96-130.57 \\ ∆\mathrm{S}°=20.07\mathrm{J}/\mathrm{Kmol} \end{gathered}$$

(c) Calculation of change of entropy for  Br2(g)+H2(g)→2HBr(g)

It is given that,

$$\begin{gathered} \mathrm{S}°({\mathrm{H}}_2,\mathrm{g})=130.57{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{Br}}_2,\mathrm{g})=245.35{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°(\mathrm{HBr},\mathrm{g})=198.59\mathrm{JK}{\mathrm{mol}}^{-1} \end{gathered}$$

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

Calculation of change of entropy is shown below:

$$\begin{gathered} ∆\mathrm{S}°=2\times \mathrm{S}°(\mathrm{HBr},\mathrm{g})-\mathrm{S}°({\mathrm{Br}}_2,\mathrm{g})-\mathrm{S}°({\mathrm{H}}_2,\mathrm{g}) \\ ∆\mathrm{S}°=2\times 198.59-245.35-130.57 \\ ∆\mathrm{S}°=21.26\mathrm{J}/\mathrm{Kmol} \end{gathered}$$

(d) Calculation of change of entropy for I(2g)+H2(g)→2HI(g)

It is given that,

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

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

Calculation of change of entropy is shown below:

$$\begin{gathered} ∆\mathrm{S}°=2\times \mathrm{S}°\left(\mathrm{HI},\mathrm{g}\right)-\mathrm{S}°({\mathrm{l}}_2,\mathrm{g})-\mathrm{S}°({\mathrm{H}}_2,\mathrm{g}) \\ ∆\mathrm{S}°=2\times 206.48-260.58-130.57 \\ ∆\mathrm{S}°21.81\mathrm{J}/\mathrm{Kmol} \end{gathered}$$