Question

(a) Calculate the change in enthalpy when $\mathbf{2}\mathbf{.}\mathbf{00}$ moles of argon is heated from$\mathbf{298}\mathbf{}\mathit{K}$ to$\mathbf{573}\mathbf{}\mathit{K}$ at constant pressure of 1 atmosphere.

(b) Calculate the change in enthalpy when moles of ethylene is taken through the same process. In both cases assume the heat capacity values predicted by equipartition are valid through the temperature range stated.

Short answer

The value of change in enthalpy is given as:

For Argon, $∆H=11.4KJ$

For, Ethylene,$∆H=73.16KJ$

Step-by-step solution

Given data

The number of moles in argon is, $n\left(Ar\right)=2.00mol$.

The initial temperature,${\text{T}}_{\text{1}}=298\text{K}$.

The final temperature,$T_2=573K$.

Pressure, $\text{p}=1\text{atm}$.

Concept of the equipartition theorem

The basis is that the energy is shared equally among all of its forms.

It gives the total average kinetic and potential energies for a system at a given temperature, from which the system's heat capacity can be computed.

Expression of specific heat at constant pressure and volume

Two specific heats are defined for gases, one for constant volume$\left(C_v\right)$and one for constant pressure$\left(C_p\right)$.

The relation between the two is obtained from the first law of thermodynamics$C_p=C_v+R$.

The general rule for determine $\left(C_v\right)$,$C_v=\frac{f}{2}R$ .

Where f is the number of degrees of freedom.

Calculation of heat capacity of a mono atomic gas

For a monoatomic ideal gas, the internal energy is all in the form of kinetic energy. For a monoatomic gases, there are three translational degrees of freedom.

$$\begin{gathered} C_v=\frac{3}{2}R \\ =1.5\times 8.314 \text{J/Kmol} \\ =12.471 \text{J/Kmol} \end{gathered}$$

Therefore, it’s $C_p$would be:

$$\begin{gathered} C_p=\frac{3}{2}R+R \\ =\frac{5}{2}R \\ =20.758 \text{J/Kmol} \end{gathered}$$

Expression of di-atomic molecule

For diatomic molecules, two rotational degrees of freedom are added, corresponding to the rotation about two perpendicular axes through the center of the molecule.

That gives us 5 total degrees of freedom including the three translational degrees of freedom.

This means that:

$$\begin{gathered} C_v=\frac{5}{2}R \\ =20.758 \text{J/Kmol} \\ {C}_p=\frac{5}{2}R+R \\ {C}_p=\frac{7}{2}R \\ =29.099 \text{J/Kmol} \end{gathered}$$

Calculation of enthalpy of Argon

Now the value of heat capacity is known. Now, calculate the enthalpy change with the help of formula:

$∆H=n{C}_p∆T$.

Put the value of given data in above equation.

$$\begin{gathered} ∆H=2.00mol\times 20.79J/Kmol\times \left(573-298\right)K \\ ∆H=11.4KJ \end{gathered}$$

Calculation of heat capacity

(i)For a polyatomic molecules having N number of atoms, there are a total of 3N degrees of freedom. Each molecule has 3 translational degrees of freedom. Linear molecule has 2 rotational degrees of freedom because the moment of inertia due to rotation around the internuclear axis is insignificant. A non-linear molecule has 3 rotational degree of freedom. Therefore, the number of vibrational degrees of freedom are :

$3N-6$For non-linear molecules

$3N-5$For linear molecules

The contribution for each vibrational degree of freedom is R. Therefore,

${\text{C}}_{\text{v}}\text{=}\frac{\text{3}}{\text{2}}\text{R +}\frac{\text{2}}{\text{2}}\text{R +}\left(\text{3N - 5}\right)\text{R}$for linear molecules.

${\text{C}}_{\text{v}}\text{=}\frac{\text{3}}{\text{2}}\text{R +}\frac{3}{\text{2}}\text{R +}\left(\text{3N - 6}\right)\text{R}$ for non-linear molecules.

Calculation of  Cp for C2H4

It is known that it is a non-linear molecule.

The structure of ethylene is shown below.

It can be seen that it is non-linear molecule.

Number of atoms: 6

1. Number of translational degrees of freedom$:\mathbf{3}\to \frac{3}{2}R$
2. Number of rotational degrees of freedom:$3\to \frac{3}{2}R$
3. Number of vibrational degrees of freedom:$=(3\times 6-6)=12 \text{R}$

So, the expression for$C_v$would be:

$C_v=\frac{3}{2}R+\frac{3}{2}R+12R$

Therefore,$C_v=15R$.

The expression for$C_p$:

$$\begin{gathered} C_p=C_v+R \\ =15R+R \\ =16R \\ =133.024\text{J}/\text{molK} \end{gathered}$$

Calculation of enthalpy change for ethylene

Calculation of enthalpy change is shown as:

$$\begin{gathered} \Delta H=2.00\times 133.024\times (573-298) \\ \Delta H=73.16\text{kJ} \end{gathered}$$