Question

(a) Calculate the change in enthalpy when $\mathbf{20}\mathbf{.}\mathbf{0}$ grams of aluminum metal are heated from $\mathbf{298}\mathbf{}\mathit{K}$ to $\mathbf{573}\mathbf{}\mathit{K}$ at constant pressure of 1 atm .

(b) Calculate the change in enthalpy when $\mathbf{20}\mathbf{.}\mathbf{0}$grams of metallic lead are 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

(a) The enthalpy change in aluminum metal is $4.96\text{kJ}$.

(b) The enthalpy change of metallic lead is $701.7\text{J}$.

Step-by-step solution

Given data

The mass of aluminum, $\text{m(Al)}=20.0\text{g}$.

The initial temperature, $T_1=298K$.

The final temperature, $T_2=573K$.

The molar mass of aluminum, $m\left(Al\right)=26.98g/mol$.

The molar mass of lead, $m\left(Pb\right)=20.0g$.

Concept of the heat capacity

The amount of heat required to change the temperature of a substance by one unit is known as the heat capacity of that substance.

Calculation of the number of moles

The number of moles for aluminum can be determined by the formula.

$\text{n =}\frac{\text{m}}{\text{M}}$ …... (i)

where, m is mass and M is the molar mass of the element.

Put the value of given data in the above formula.

$$\begin{gathered} n\left(Al\right)=\frac{20.0\text{g}}{26.98{\text{gmol}}^{-1}} \\ =0.741\text{mol} \end{gathered}$$

(a) Calculation of change in enthalpy (Al)

Change in enthalpy can be determined with the help formula.

$\Delta H=n{c}_{\text{p}}\Delta T$ …. (ii)

Where, $\Delta H$is enthalpy change, n is number of moles, $\Delta T$is a change in temperature, $c_p$is$24.35{\text{JK}}^{-1}{\text{mol}}^{-1}$.

Substitute the value of given data in equation (ii).

$$\begin{gathered} \Delta H=0.741\text{mol}\times 24.35{\text{JK}}^{-1}{\text{mol}}^{-1}\times (573-298)\text{K} \\ =0.741\text{mol}\times 24.35{\text{JK}}^{-1}{\text{mol}}^{-1}\times 275\text{K} \\ =4.96\times {10}^3\text{J} \end{gathered}$$

Therefore, the enthalpy change,$\Delta H=4.96kJ$ .

(b) Calculation of change of enthalpy for lead (Pb) 

First, determine the number of moles of lead atom with the help of equation (i).

$$\begin{gathered} n\left(Pb\right)=\frac{20.0\text{g}}{207.2{\text{gmol}}^{-1}} \\ =0.096\text{mol} \end{gathered}$$

Now, use equation (ii) to calculate enthalpy change for lead.

It is given that,${\text{C}}_{\text{p}}=26.44{\text{JK}}^{-1}{\text{mol}}^{-1}$ for Pb .Now, put the value in equation (ii).

$$\begin{gathered} \Delta H=0.0965\text{mol}\times 26.44{\text{Jmol}}^{-1}{\text{K}}^{-1}\times (573-298)\text{K} \\ =0.0965\text{mol}\times 26.44{\text{Jmol}}^{-1}{\text{K}}^{-1}\times 275\text{K} \\ =701.7\text{J} \end{gathered}$$

Therefore, enthalpy change for lead is $701.7J$.