Question

For a reaction, ${\rm{A}}\left( {\rm{g}} \right) \to {\rm{A}}\left( {\rm{l}} \right){\rm{;\Delta H}} = - {\rm{3RT}}$. The correct statement for the reaction is:
A. $\left| {{\rm{\Delta H}}} \right| < \left| {{\rm{\Delta U}}} \right|$
B. $\left| {{\rm{\Delta H}}} \right| > \left| {{\rm{\Delta U}}} \right|$
C. ${\rm{\Delta H}} = {\rm{\Delta U}} = {\rm{O}}$
D. ${\rm{\Delta H}} = {\rm{\Delta U}} \ne {\rm{O}}$

Answer 1

We know that the enthalpy and internal energy both are different. The internal energy is that which is generally used to create or destroy the system.

Formula used: ${\rm{\Delta H}} = {\rm{\Delta U}} + \left( {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right){\rm{RT}}......\left( 1 \right)$
Where, delta H is the change in enthalpy, delta U is the change in internal energy, delta n is the change in the moles of gas, R is the gas constant and T is the temperature of gas.

Complete step-by-step solution: As we know that, the enthalpy can be shown by the symbol H and internal energy can be shown by symbol U.
The relation between enthalpy and internal energy can be shown by using the equation is depicted below:
${\rm{\Delta H}} = {\rm{\Delta U}} + \left( {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right){\rm{RT}}......\left( 1 \right)$
Where, delta H is the change in enthalpy, delta U is the change in internal energy, delta n is the change in the moles of gas, R is the gas constant and T is the temperature of gas.
The value of $\left( {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right)$ is the subtraction of moles of products and the moles of reactant of the gas.
So, the value of the product is zero while the value of reactant is one.
Thus, $\left( {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right) = 0 - 1\\\ \left( {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right) = - 1$
The given value of enthalpy is ${\rm{\Delta H}} = - {\rm{3RT}}$.
Substitute the respective values in equation 1 we get,
$\Rightarrow - 3{\rm{RT}} = {\rm{\Delta U}} + \left( { - 1} \right){\rm{RT}}\\\ \Rightarrow {\rm{\Delta U}} = - 3{\rm{RT}} + {\rm{RT}}\\\ \therefore {\rm{\Delta U}} {\rm{ = }} - 2{\rm{RT}}$
So, the obtained value of change in internal energy is ${\rm{\Delta U}} = - 2{\rm{RT}}$.
So, the value of change enthalpy is greater than the value of change in internal energy and the expression is shown as ${\rm{\Delta H}} > \Delta {\rm{U}}$.

Therefore, the correct option for this question is B that is $\left| {{\rm{\Delta H}}} \right| > \left| {{\rm{\Delta U}}} \right|$.

Note: Thermodynamics is one of the main branches of chemistry. The enthalpy, entropy and inter energy all come under the category of thermodynamics. The work done is also part of thermodynamics.