Question

For the reaction CO(g) + \(\frac{1}{2}\) O\(_2\)(g) \(\rightarrow\) CO\(_2\)(g) Which one of the statement is correct at constant T and P?

• A. \(\Delta H = \Delta E\)
• B. \(\Delta H \textless \Delta E\)
• C. \(\Delta H \textgreater \Delta E\)
• D. \(\Delta H\) is independent of physical state of the reactants

Hint:

Just count the moles of gases!
If products have fewer moles of gas (\(\Delta n_g \textless 0\)), then \(\Delta H \textless \Delta E\).
If products have more moles of gas (\(\Delta n_g \textgreater 0\)), then \(\Delta H \textgreater \Delta E\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta E$, also denoted as $\Delta U$) for a gaseous reaction is governed by the change in the number of moles of gaseous products and reactants.
Step 2: Key Formula or Approach:
\[ \Delta H = \Delta E + \Delta n_g RT \] where $\Delta n_g = (\text{Total moles of gaseous products}) - (\text{Total moles of gaseous reactants})$.
Step 3: Detailed Explanation:
1. Write the balanced equation: $CO(g) + \frac{1}{2}O_{2}(g) \rightarrow CO_{2}(g)$.
2. Calculate $\Delta n_g$: - Moles of products = 1 ($CO_2$). - Moles of reactants = $1 + 0.5 = 1.5$ ($CO$ and $O_2$). - $\Delta n_g = 1 - 1.5 = -0.5$.
3. Substitute $\Delta n_g$ into the formula: \[ \Delta H = \Delta E + (-0.5)RT \] \[ \Delta H = \Delta E - 0.5RT \]
4. Since $R$ and $T$ are positive, we are subtracting a positive value from $\Delta E$ to get $\Delta H$. This means $\Delta H$ is smaller than $\Delta E$.
Step 4: Final Answer
At constant $T$ and $P$, $\Delta H<\Delta E$.