Question

What is the difference between the enthalpy change and internal energy change for the formation of one mole of ammonia in standard state?

• A. $-\frac{1}{2}RT$
• B. $-2RT$
• C. $-\frac{3}{2}RT$
• D. $-RT$

Answer 1

Answer: (D)

-RT

Answer 2

For the formation of 1 mole of ammonia, the reaction is:

$$\frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g)$$

1. Calculate the change in the number of moles of gas:

   • Moles of gas in reactants: $\frac{1}{2} + \frac{3}{2} = 2$
   • Moles of gas in product: $1$
   • $\Delta n_{\text{gas}} = 1 - 2 = -1$

2. Use the relation:

$$\Delta H = \Delta U + \Delta n_{\text{gas}} \cdot RT$$

Therefore,

$$\Delta H - \Delta U = -RT$$