Question

What is the value of \(\Delta H - \Delta U\) for the formation of 2 moles of ammonia from \(H_{2(g)}\) and \(N_{2(g)}\)?

• A. \(-\frac{RT}{2}\)
• B. \(\frac{RT}{2}\)
• C. \(-2RT\)
• D. \(2RT\)

Hint:

\(\Delta H = \Delta U + \Delta n_g RT\). For reactions where number of gas moles decreases, \(\Delta H < \Delta U\) (negative difference).

Answer 1

The Correct Option is C

Step 1: Understand the question.
We must find $\Delta H - \Delta U$ for making 2 moles of ammonia from nitrogen and hydrogen gas.
Step 2: Recall the linking formula.
Enthalpy and internal energy differ by the gas work term: \[ \Delta H - \Delta U = \Delta n_g RT \] Here $\Delta n_g$ is the change in moles of gas.
Step 3: Write the balanced reaction.
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
Step 4: Count the gas moles.
Products have 2 moles of gas. Reactants have $1 + 3 = 4$ moles of gas.
Step 5: Find the change in gas moles.
\[ \Delta n_g = 2 - 4 = -2 \] So $\Delta H - \Delta U = -2RT$.
Step 6: Choose the answer.
The value is $-2RT$, which is option 3. \[ \boxed{-2RT} \]