Question

128. For rubidium $\mathbf{∆}{\mathbf{H}}_{\mathbf{vap}}^{\mathbf{o}}\mathbf{=}\mathbf{69}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$at 686°C, its boilingpoint. Calculate $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}\mathbf{,}\mathbf{q}\mathbf{,}\mathbf{w}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{E}$ for the vaporization of 1.00 mole of rubidium at 686°C and 1.00 atm pressures.

Short answer

The values of vaporization of 1.00 mole of rubidium at 686°C and 1.00 atm pressures are,

$$\begin{gathered} \mathbf{q}=∆{\mathrm{H}}_{\mathrm{vap}}^{\mathrm{o}}=69.0\mathrm{kJ}/\mathrm{mol} \\ \mathbf{w}=-7973\mathrm{J} \\ \mathbf{∆}\mathbf{E}=61.0\mathrm{kJ}/\mathrm{mol} \\ \mathbf{∆}\mathbf{S}=71.9\mathrm{J}/\mathrm{mol}.\mathrm{K} \end{gathered}$$

Step-by-step solution

Step 1: Introduction to the Concept

Thework done is computed as follows:
$\mathbf{w}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{V}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$
Here,

w = work done
P =pressure
V =volume

Theinternal energy is computed as follows:
$\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{q}\mathbf{+}\mathbf{w}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(2\right)$
Here,
$\mathbf{∆}\mathbf{E}=$energy change
q =heat
w = work done

Step 2: Determination of ∆So,q,w 

From the given,

$∆{\mathrm{H}}_{\mathrm{vap}}^{\mathrm{o}}=69.0\mathrm{kJ}/\mathrm{mol}$at a boiling point 686°C and $\mathrm{n}=1.00\mathrm{mol}$

Now,

$\begin{array}{rcl}\mathbf{q}& \mathbf{=}& \mathbf{∆}{\mathbf{H}}_{\mathbf{vap}}^{\mathbf{o}}\\ & =& 69.0\mathrm{kJ}/\mathrm{mol}\end{array}$

From equation (1),

$\begin{array}{rcl}\mathbf{w}& \mathbf{=}& \mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{V}\mathbf{=}\mathbf{-}\mathbf{∆}\mathbf{nRT}\\ & =& -1.00\mathrm{mol}\times 8.314\mathrm{J}/\mathrm{Kmol}\times 959\mathrm{K}\\ \mathrm{w}& =& -7973\mathrm{J}\\ & =& -7.9\mathrm{kJ}\\ & & \end{array}$

Step 3: Determination of ∆E 

From equation (2),

$\begin{array}{rcl}\mathbf{∆}\mathbf{E}& \mathbf{=}& \mathbf{q}\mathbf{+}\mathbf{w}\\ & =& 69.0\mathrm{kJ}/\mathrm{mol}-7.97\mathrm{kJ}/\mathrm{mol}\\ & =& 61.0\mathrm{kJ}/\mathrm{mol}\end{array}$

$\begin{array}{rcl}\mathbf{∆}\mathbf{S}& \mathbf{=}& \frac{\mathbf{∆}{\mathbf{H}}_{\mathbf{vap}}}{{\mathbf{T}}_{\mathbf{bp}}}\\ & =& \frac{69000\mathrm{J}/\mathrm{mol}}{9.59\mathrm{K}}\\ & =& 71.9\mathrm{J}/\mathrm{molK}\end{array}$

Therefore

$$\begin{gathered} \mathbf{q}=∆{\mathrm{H}}_{\mathrm{vap}}^{\mathrm{o}}=69.0\mathrm{kJ}/\mathrm{mol} \\ \mathbf{w}=-7973\mathrm{J} \\ \mathbf{∆}\mathbf{E}=61.0\mathrm{kJ}/\mathrm{mol} \\ \mathbf{∆}\mathbf{S}=71.9\mathrm{J}/\mathrm{mol}.\mathrm{K} \end{gathered}$$