Question

If standard molar enthalpy change and standard molar internal energy change measured in bomb calorimeter are equal, which one of the following statements is correct?

• A. $\Delta n>0$, with increase in pressure
• B. $\Delta n>0$, with decrease in pressure
• C. $\Delta n<0$, with increase in pressure
• D. $\Delta n = 0$, at constant pressure

Hint:

The difference between $\Delta H$ and $\Delta U$ is significant only when there is a change in the number of moles of gaseous species during a reaction.

Answer 1

The Correct Option is D

Step 1: Recall the relationship between $\Delta H$ and $\Delta U$.
$\Delta H = \Delta U + \Delta (PV) = \Delta U + RT \Delta n_g$
Step 2: Apply the condition $\Delta H = \Delta U$.
If $\Delta H = \Delta U$, then $RT \Delta n_g = 0$.
Since $R$ and $T$ are generally non-zero, $\Delta n_g = 0$.
Step 3: Interpret $\Delta n_g = 0$.
$\Delta n_g = 0$ means there is no change in the number of moles of gaseous species in the reaction.
Step 4: Evaluate the options.
The equality $\Delta H = \Delta U$ implies $\Delta n_g = 0$. Option 4 states $\Delta n = 0$, which refers to the change in the number of moles of gaseous species. The condition of constant pressure is typical for measuring enthalpy changes, and the equality holds when $\Delta n_g = 0$.
Thus, the correct statement is $ \boxed{\Delta n = 0, at constant pressure} $.