Question

The value of enthalpy change ($\Delta H$) for the reaction \[{C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)\], at $27^\circ C$ is \[ - 1366.5{{ }}kJ\;mo{l^{ - 1}}\]. The value of internal energy change for the above reaction at this temperature will be:
A. $ - 1371.5kJ$
B. $ - 1369.0kJ$
C. $ - 1364.0kJ$
D. $ - 1361.5kJ$

Answer 1

Hint:As the enthalpy change is given to us and the temperature doesn’t change, by assuming that the reaction occurs at constant pressure we can use the ideal gas relation in the equation relating enthalpy and internal energy.
Formulas used:
$\Delta H = \Delta U + P\Delta V$
Where $\Delta H$ is the enthalpy change, $\Delta U$ is the change in internal energy, $P$ is the pressure, $\Delta V$ is the change in volume.
$PV = nRT$
Where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant and $T$ is the absolute temperature.
$\Delta H = \Delta U + \Delta {n_g}RT$
Where $\Delta {n_g}$ is the change in number of moles in gaseous phase, and the rest of the terms are the same as that defined above.

Complete step by step answer:
As we know, enthalpy and internal energy are related by the equation:
$\Delta H = \Delta U + P\Delta V$
Where $\Delta H$ is the enthalpy change, $\Delta U$ is the change in internal energy, $P$ is the pressure, $\Delta V$ is the change in volume.
From the ideal gas equation, we have:
$PV = nRT$
Where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant and $T$ is the absolute temperature.
In this case, as pressure is assumed to be constant and so is the temperature, the only quantities which change before and after the reaction are volume and number of moles. Therefore, the equation reduces to:
$P\Delta V = \Delta {n_g}RT$
Where, $\Delta V$ is the change in volume and $\Delta {n_g}$ is the change in number of moles in gaseous phase
Substituting this in our equation for enthalpy, we get:
$\Delta H = \Delta U + \Delta {n_g}RT \Rightarrow \Delta U = \Delta H - \Delta {n_g}RT$
In this reaction, there are two components present in the gaseous phase, namely, oxygen and carbon dioxide, on the reactant and product side respectively. Three moles of oxygen are present in the reactant side while one mole of carbon dioxide is present in the product side.
Therefore, the change in number of moles in gaseous phase:
$\Delta {n_g} = 2 - 3 = - 1$
We know $R = 8.314J/mol.K$ and by converting temperature from degree Celsius to Kelvin by adding 273, we have:
$T = 27 + 273 = 300K$
Substituting these values into the equation $\Delta U = \Delta H - \Delta {n_g}RT$, we have:
$\Delta U = - 1366.5 + ( - 1 \times 8.314 \times 300)$
On solving, we get:
$\Delta U = - 1364000J = - 1364.0kJ$
Hence, the correct option is C.


Note:
Note that we are only required to calculate the change in number of moles in the gaseous phase, and not for the entire reaction.
Also the equations we derived here are a direct consequence of the First Law of Thermodynamics, which states that the heat supplied/released in a reaction is equal to the sum of change in internal energy and work done by/on the gas. In this case, as the temperature and pressure are kept constant, the heat released is equal to the enthalpy change.