Question

For which change \[\Delta H \ne \Delta U\]
A. \[{H_{2(g)}} + {I_{2(g)}} \rightleftharpoons 2H{I_{(g)}}\]
B. \[HC{l_{(l)}} + NaO{H_{(l)}} \to NaC{l_{(l)}} + {H_2}{O_{(l)}}\]
C. \[{C_{(s)}} + {O_{2(g)}} \rightleftharpoons C{O_{2(g)}}\]
D. \[{N_{2(g)}} + 3{H_{2(g)}} \to 2N{H_{3(g)}}\]

Answer 1

Hint: To solve this question, we must first derive a relation between \[\Delta H\] or change in enthalpy and \[\Delta U\] or change in internal energy. After this, we must identify the conditions which lead to the values of these two quantities being equal. We must then eliminate the reactions that abide by these conditions.

Complete step by step answer:
Before we move ahead with the solution of this question, let us understand a few basic concepts.
 \[\Delta H\] or change in enthalpy: This quantity accounts for the change in the total energy in the given reaction. To put it in simpler terms, it can be understood as the change in the energy states observed at two different instances in the reaction.
 \[\Delta U\] or change in internal energy: Internal energy of a system can be identified as a thermodynamic property which represents the sum of the total kinetic energy and potential energy of the system. These kinetic and potential energies are associated with the random motion of the molecules present in the object. The internal energy of a sample is affected by the temperature of the system.
The relation between the enthalpy of a reaction and the internal energy of the system can be represented by the following formula:
 \[\Delta H = \Delta U + \Delta nRT\]
Where n represents the number of moles of the substance, T is the temperature of the system and R is the gas constant. These factors directly affect the values of the enthalpy as well as the internal energy of the system. The values of enthalpy and internal energy would not be equal to each other until and unless either the change in the number of moles or the temperature is zero. In the options given, let us consider that these reactions are carried out at constant temperature.
1. \[{H_{2(g)}} + {I_{2(g)}} \rightleftharpoons 2H{I_{(g)}}\]
 \[\Delta n = 2 - 2 = 0\]
2. \[HC{l_{(l)}} + NaO{H_{(l)}} \to NaC{l_{(l)}} + {H_2}{O_{(l)}}\]
 \[\Delta n = 2 - 2 = 0\]
3. \[{C_{(s)}} + {O_{2(g)}} \rightleftharpoons C{O_{2(g)}}\]
 \[\Delta n = 2 - 2 = 0\]
4. \[{N_{2(g)}} + 3{H_{2(g)}} \to 2N{H_{3(g)}}\]
 \[\Delta n = 4 - 2 = 2\]
Hence, \[\Delta H \ne \Delta U\] only in the last reaction.

Hence, Option D is the correct option.

Note:
The equation used in the solution is derived from the first law of thermodynamics which states that change observed in the enthalpy of the reaction is equal to the value obtained by subtracting work done from the net heat transfer (in this case internal energy).