Question

The normal boiling point of liquid ammonia is$\mathbf{240}\mathbf{}\mathbf{\text{K}}$ ; the enthalpy of vaporization at that temperature is$\mathbf{23}\mathbf{.}\mathbf{4}\mathbf{}{\mathbf{\text{kJmol}}}^{\mathbf{-}\mathbf{1}}$ . The heat capacity of gaseous ammonia at constant pressure is$\mathbf{38}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{-}\mathbf{1}}$ .

(a) Calculate$\mathit{q}\mathbf{,}\mathit{w}\mathbf{,}\mathit{\Delta }\mathit{H}$ , and $\mathbf{∆}\mathit{U}$for the following change in state:

$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{\text{molNH}}}_{\mathbf{3}}\mathbf{(}\mathit{\ell }\mathbf{,}\mathbf{1}\mathbf{}\mathbf{\text{atm}}\mathbf{,}\mathbf{240}\mathbf{}\mathbf{\text{K}}\mathbf{)}\mathbf{\to }\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{\text{molNH}}}_{\mathbf{3}}\mathbf{(}\mathit{g}\mathbf{,}\mathbf{1}\mathbf{}\mathbf{\text{atm}}\mathbf{,}\mathbf{298}\mathbf{}\mathbf{\text{K}}\mathbf{)}$

Assume that the gas behaves ideally and that the volume occupied by the liquid is negligible.

(b) Calculate the entropy of vaporization of ${\mathbf{\text{NH}}}_{\mathbf{3}}$at$\mathbf{240}\mathbf{}\mathbf{\text{K}}$ .

Short answer

The$q$ is$46.8\text{kJ}$ , $W$is$-4.91\text{kJ}$ , $\Delta H$is $51.76\text{kJ}$, $∆U$is $41.89\text{kJ}$ and entropy of vaporization$\Delta S_{vap}$ is$195\text{J}{\text{K}}^{-1}$ .

Step-by-step solution

Given information

The process of vaporization of${\text{NH}}_3$is observed as:

${\text{NH}}_3\left(l\right)\to {\text{NH}}_3\left(g\right)$

The physical properties of reaction conditionare:

$T_B=240\text{K,}\Delta H_{vap}=23.4\text{kJ}{\text{mol}}^{-1}and{C}_p=38\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$

Were,$n=2.00\text{mol}{T}_1=240\text{K}{T}_2=298\text{K}{p}_1=p_2=1.00\text{atm}$

And assuming that the gas behaves ideally and that the volume occupied by the liquid is negligible$V_1=0$ .

Concept of entropy and enthalpy of vaporization

The degree of unpredictability in the vapor and liquid phases is related to the entropy of vaporization. It refers to the translational, rotational, and conformational motion of molecules in a pure component. The translational impact is the most important contributor to vaporization entropy.

The enthalpy of vaporization also known as the (latent) heat of vaporization or heat of evaporation, is the amount of energy (enthalpy) required to convert a quantity of a liquid substance into a gas. The enthalpy of vaporization depends on the pressure at which the transformation occurs

Calculate the enthalpy change

$$\begin{gathered} \Delta H=\Delta q+\Delta \left(pV\right)\backslash \mathrm{hfill} \\ \Delta H=\Delta q+V\Delta p+p\Delta V\backslash \mathrm{hfill} \end{gathered}$$

Since the pressure stays constant in our reaction, the equation will look like this:

$\Delta H=\Delta q+p\Delta V$

Calculate the heat exchanged

The heat exchanged is really the heat necessary for the vaporization to take place. Therefore,it can be written as:$q=n\Delta H_{vap}$ .

Calculate the heat

Given values.$n=2.00\text{mol}\Delta H_{vap}=23.4\text{kJ}{\text{mol}}^{-1}$

Substitute them:

$$\begin{gathered} q=2.00\times 23.4 \\ q=46.8\text{kJ}{\text{mol}}^{-1} \end{gathered}$$

The formula for the work done in this process is$w=-p\Delta V$

Since the initial volume is negligible, $\Delta V=V_2-0=V_2$

Step 6:Calculate the final volume

Thus, by using the ideal gas law:

$$\begin{gathered} pV=nRT \\ V=\frac{nRT}{p} \end{gathered}$$

Where $n=2.00\text{mol}T=298\text{K}p=1.00\text{atm}$

Therefore, the equation is:

$$\begin{gathered} V_2=\frac{2.00\times 0.08206\times 298}{1.00} \\ {V}_2=48.91\text{L} \end{gathered}$$

Calculate the work done

To get the value in J, use Pascals instead of atmospheres$1\text{atm}=101325\text{Pa}$.

There is also need to express the volume in${\text{m}}^3$.Then put the information into the formula:

$$\begin{gathered} w=-101325\times 48.91\times {10}^{-3} \\ w=4955.81\text{J} \\ w=-4.91\text{kJ} \end{gathered}$$

Calculate the enthalpy change

With all the necessary information, proceed ahead and put it into our formula:

$$\begin{gathered} \Delta H=46.8+101.325\times 48.91 \\ \Delta H=51.76\text{kj} \end{gathered}$$

Calculate the internal energy change

The formula goes$\Delta U=q-p\Delta V$

Substitute the values:

$$\begin{gathered} \Delta U=46.8-4.91 \\ \Delta U=41.89\text{kJ} \end{gathered}$$

Calculate the entropy change of vaporization for this reaction.

The formula for the change in entropy of vaporization is$\Delta S_{vap}=?$

$$\begin{gathered} \Delta S_{vap}=\frac{2.00\times 23.4\times {10}^3}{240} \\ \Delta S_{vap}=195\text{J}{\text{K}}^{-1} \end{gathered}$$