Question

For a reaction, $ {A_(g) \to A_(l); \Delta H = - 3RT}$. The correct statement for the reaction is :

• A. $\Delta H = \Delta U \neq O $
• B. $\Delta H = \Delta U = O $
• C. $ | \Delta H| < | \Delta U|$
• D. $ | \Delta H| > | \Delta U|$

Answer 1

The Correct Option is D

To analyze the given reaction $ {A_(g) \to A_(l); \Delta H = - 3RT}$, we need to understand the relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) under different conditions. The relation between \(\Delta H\) and \(\Delta U\) is given by:

\[\Delta H = \Delta U + \Delta nRT\]

where \(\Delta n\) is the change in the number of moles of gas.

1. This reaction converts a gaseous substance \(A_{(g)}\) to its liquid form \(A_{(l)}\). Hence, there is a decrease in the number of moles of gas from 1 to 0, making \(\Delta n = -1\).
2. Substitute \(\Delta n = -1\) into the equation:

\[\Delta H = \Delta U + (-1)RT = \Delta U - RT\]

3. Given that \(\Delta H = -3RT\), we can set up the equation:

-3RT = \Delta U - RT

Simplifying this, we solve for \(\Delta U\):

\[\Delta U = -3RT + RT = -2RT\]

4. Now let's compare the magnitudes of \(\Delta H\) and \(\Delta U\):

The absolute values are:

| \Delta H | = 3RT

| \Delta U | = 2RT

Since 3RT > 2RT, it follows that:

| \Delta H | > | \Delta U |

Therefore, the correct statement is: $ | \Delta H| > | \Delta U|$.

Hence, the correct option is $ | \Delta H| > | \Delta U|$ and we have correctly demonstrated the reasoning behind this conclusion using the principles of thermodynamics.