Q12.20P - Chemistry: Molecular Nature Of Matter And Change

Question

From the data below, calculate the total heat (in J) needed to convert 0.333 mole of gaseous ethanol at ${300}^o\mathbf{C}$   and 1 atm to liquid ethanol at $25{.0}^{\mathrm{o}}\mathrm{C}$ and 1 atm: B.P. at 1 atm: , $78{.5}^o\mathbf{C}$;${\mathbf{\Delta H}}_{\mathrm{yap}}$: 40.5 kJ/mol; : ${\mathbf{C}}_{\mathrm{gas}}$1.43 J/ ${\mathbf{g}}^o\mathbf{C}$; : 2.45 J/${\mathbf{g}}^o\mathbf{C}$.

Step-by-Step Solution

Answer

The heat required in the process is 20356.06 J

Step1: Heat needed to convert 0.333 mol liquid at 25.0 o C to 78.5 o C

The heat required to convert 0.333 mol (15.34 g) liquid at $25.0^{o}C$to $78.5^{o}C$is calculated as:

$\Delta H_1=C_{solid}(J/g^{o}C)*Weight\left(g\right)*\Delta T{(}^{o}C)$

$\Delta H_1=2.45*15.34*53.5=2010.69J$

Step2: Heat needed to convert 0.333 mol liquid to vapour

The heat required to convert 0.333 mol liquid to vapour is 

 $\begin{array}{c}\Delta H_2=40500J/mol\\ =40500*0.333J\\ \Delta H_2=13486.5J\end{array}$

Step3: Heat needed to convert 0.333 mol liquid at 78.5 o C to 300 o C

The heat required to convert 0.333 mol (15.34 g) liquid at $78.5^{o}C$to${300}^{o}C$ is calculated as:

$\Delta H_3=C_{solid}(J/g^{o}C)*Weight\left(g\right)*\Delta T{(}^{o}C)$

Hence, the total heat energy required in the process is 

$\begin{array}{c}\Delta H=\Delta H_1+\Delta H_2+\Delta H_3\\ =2010.69+13486.5+4858.87\\ \Delta H=20356.06J\end{array}$