Question

What will be $\Delta H$ for the reaction; $Ag\left( s \right) + \dfrac{1}{2}H{g_2}C{l_2}\left( s \right) \to AgCl\left( s \right) + Hg\left( l \right)$ at ${25^ \circ }C$, if this reaction can be conducted in a cell for which the $emf = 0.0455\;volt$ at this temperature with temperature coefficient $3.389 \times {10^{ - 4}}\;volt\,{\deg ^{ - 1}}$?
A.$ + 1280\,cal$
B.$ + 640\,cal$
C.$ - 1280\;cal$
D.$ - 640\,cal$

Answer 1

Hint: The measurement of energy in a thermodynamic system is termed as Enthalpy. The total content of heat of a system and equivalent to the internal energy of the system plus the product of volume and pressure is equal to the amount of enthalpy.

Complete answer:
Basically, the internal energy which is required to generate a system is known as enthalpy. The quantity of energy which is required to make room for it by organizing its pressure and volume and displacing its environment.
It is important to remember that any system comprises multiple participants, each participant has its pressure and volume and we know that the product of pressure and volume of the specific system is constant. This enthalpy is equivalent to the summation of internal energy of such a system with the constant.
As like other scientific theories, enthalpy also possesses a mathematical formula. The equation of enthalpy is given below:
$H = U + PV$
And the equation of change in enthalpy is shown below:
$\Delta H = \Delta U + \Delta PV$
In the above equation the volume and temperature don’t change throughout the process.
When there is no change in pressure and temperature throughout the process and the task restricted to pressure and volume, the change in enthalpy is:
$\Delta H = \Delta U + P\Delta V$
According to the question:
$Ag\left( s \right) + \dfrac{1}{2}H{g_2}C{l_2}\left( s \right) \to AgCl\left( s \right) + Hg\left( l \right)$
As it is given,
$E = 0.0455\,volt$
Temperature coefficient is,
$\dfrac{{dE}}{{dT}} = 3.389 \times {10^{ - 4}}\;Volt\;{\deg ^{ - 1}}$
We know that,
$ - nFE = \Delta H - nFT\left( {\dfrac{{dE}}{{dT}}} \right)$
$ \Rightarrow E = - \dfrac{{\Delta H}}{{nF}} + T\left( {\dfrac{{dE}}{{dT}}} \right)$
$ \Rightarrow 0.0455 = - \dfrac{{\Delta H}}{{1 \times 96485}} + 298 \times 3.389 \times {10^{ - 4}}$
$\Delta H = + 5354.16\;J$
$ \Rightarrow + 1280\,\,Cal$

Note:
Enthalpy is very much essential as it gives us information on how much heat is in the system. Heat is essential as by it we can derive valuable work. An enthalpy shift tells us how much enthalpy was obtained or lost in the chemical reaction.