The standard molar enthalpy of formation of NH3(g) is –45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g) react to produce NH3(g)?
A. –10.3 kJ
B. –20.7 kJ
C. –29.8 kJ
D. –43.7 kJ
E. –65.6 kJ
The Correct Answer and Explanation is :
To determine the enthalpy change when 9.51 g of N₂(g) and 1.96 g of H₂(g) react to produce NH₃(g), we need to follow these steps:
Step 1: Write the balanced chemical equation
The balanced equation for the formation of ammonia is:
[
N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)
]
Step 2: Calculate the moles of N₂ and H₂
We start by calculating the moles of N₂ and H₂ based on their given masses.
- Molar mass of N₂ = 28.0 g/mol
- Molar mass of H₂ = 2.0 g/mol
For N₂:
[
\text{moles of N}_2 = \frac{\text{mass of N}_2}{\text{molar mass of N}_2} = \frac{9.51 \, \text{g}}{28.0 \, \text{g/mol}} = 0.339 \, \text{mol}
]
For H₂:
[
\text{moles of H}_2 = \frac{\text{mass of H}_2}{\text{molar mass of H}_2} = \frac{1.96 \, \text{g}}{2.0 \, \text{g/mol}} = 0.98 \, \text{mol}
]
Step 3: Determine the limiting reactant
From the balanced equation, we see that 1 mol of N₂ reacts with 3 mol of H₂. Therefore, we need 3 moles of H₂ for every mole of N₂. Let’s check if we have enough H₂ to react with all the N₂:
- Required moles of H₂ for 0.339 mol of N₂:
[
0.339 \, \text{mol N}_2 \times 3 = 1.017 \, \text{mol H}_2
]
Since we only have 0.98 mol of H₂, which is less than the required 1.017 mol, H₂ is the limiting reactant.
Step 4: Calculate the enthalpy change
Now, we know that 1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃. The standard molar enthalpy of formation of NH₃(g) is –45.9 kJ/mol. The enthalpy change for the reaction is proportional to the amount of limiting reactant, H₂.
For every 3 moles of H₂, 2 moles of NH₃ are formed. The enthalpy change for 2 moles of NH₃ is:
[
\Delta H = 2 \times (-45.9 \, \text{kJ/mol}) = -91.8 \, \text{kJ}
]
Now, calculate the enthalpy change for the 0.98 mol of H₂:
[
\Delta H_{\text{reaction}} = \frac{0.98}{3} \times (-91.8 \, \text{kJ}) = -29.8 \, \text{kJ}
]
Thus, the enthalpy change for the reaction is –29.8 kJ.