Question

How would you calculate the enthalpy change delta H, for the process in which $33.3g$ of water is converted from liquid at $4.6C{}^{\circ }$ to vapor at $25.0C$ ? For water H= $44.0$KJ/mole at $25C$ and s= $4.18$J/g C for water (l)

Answer 1

Hint:We know that the Vaporization of water occurs at constant pressure. And when a process occurs at constant pressure, we know that $q=\Delta H$. It means that the total heat required for vaporization is equal to enthalpy change. So, to solve this question we just need to calculate total heat required for the process and it will be equal to enthalpy.

Complete step-by-step answer:In the question liquid is at temperature of $4.6$ degree Celsius. So, in order to vaporize the liquid, we need to proceed through two basic steps.
Step 1. Change of temperature of liquid from $4.6\,to\,25{{C}^{\circ }}$
Step 2. vaporization of the liquid at $25{{C}^{\circ }}$
If we calculate the heat required in both the process and add them then we can easily calculate the enthalpy.
Let us calculate the heat required for the Step1
The heating of the water occurs at constant pressure and hence the heat required is given by the relation ${{q}_{1}}=m{{C}_{p}}\Delta T$$\left( i \right)$
In the above formula, $m=$ mass of the liquid in grams
${{C}_{p}}\left( s \right)=$ heat capacity at constant pressure
$\Delta T=$ change in the temperature
In the question values of $m=$$33.3g$
${{C}_{p}}\left( s \right)=4.18$J/g C
$\Delta T=25-4.6=20.4{{C}^{\circ }}$
Substituting these values in equation $\left( i \right)$, we get
$\begin{align}
  & {{q}_{1}}=33.3g\times 4.184J/g{{C}^{\circ }}\times 20.4{{C}^{\circ }} \\
 & \therefore {{q}_{1}}=2842.27J\,or\,2.842Kj \\
\end{align}$
Hence, for step 1 we need $2842.7J$ of heat
Now let us calculate heat required for step 2

For this the heat will be given as
$\begin{align}
  & {{q}_{2}}={{n}_{w}}\Delta {{H}_{vap}} \\
 & {{q}_{2}}=\dfrac{33.3g}{18}\times 44.0Kj/mol \\
 & \therefore {{q}_{2}}=81.33kJ \\
\end{align}$
Now, the total heat for both the process and complete vaporization is given as
$\begin{align}
  & {{q}_{total}}={{q}_{1}}+{{q}_{2}} \\
 & {{q}_{total}}=2.842+81.33 \\
 & \therefore {{q}_{total}}=84.17kJ \\
\end{align}$
Now, we already know that vaporization occurs at constant pressure and if a process occurs at constant pressure then
$\begin{align}
  & {{q}_{total}}=\Delta H \\
 & \therefore \Delta H=84.17kJ \\
\end{align}$
Hence, the enthalpy required for the reaction is $84.17kJ$.

Note:enthalpy change is the measure of heat change taking place during the place during the process at constant temperature and constant pressure. ${{q}_{p}}=\Delta H$
Enthalpy $\Delta H$ and internal energy change $\Delta U$ are related as $\Delta H = \Delta U + \Delta n_g RT$