The standard molar enthalpy of formation of NH3(g) is 45.9 kJ/mol. What is the enthalpy change if 9.10 g N2(g) and 1.82 g H2(g) react to produce NH3(g)?
The Correct Answer and Explanation is :
To find the enthalpy change for the reaction where 9.10 g of N₂(g) and 1.82 g of H₂(g) react to produce NH₃(g), we first need to write the balanced chemical equation for the reaction:
[
N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)
]
Step 1: Convert the masses of N₂ and H₂ to moles
To calculate the moles of each reactant, we use the molar masses of N₂ and H₂.
- Molar mass of N₂ = 28.02 g/mol
- Molar mass of H₂ = 2.02 g/mol
For N₂:
[
\text{Moles of } N_2 = \frac{9.10 \, \text{g}}{28.02 \, \text{g/mol}} = 0.325 \, \text{mol}
]
For H₂:
[
\text{Moles of } H_2 = \frac{1.82 \, \text{g}}{2.02 \, \text{g/mol}} = 0.901 \, \text{mol}
]
Step 2: Determine the limiting reagent
From the balanced equation, we see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, we can compare the mole ratios to identify the limiting reagent.
- For 0.325 mol of N₂, we would need ( 0.325 \, \text{mol} \times 3 = 0.975 \, \text{mol} ) of H₂.
- We only have 0.901 mol of H₂, which is less than 0.975 mol. Thus, H₂ is the limiting reagent.
Step 3: Calculate the moles of NH₃ produced
From the balanced equation, 3 moles of H₂ produce 2 moles of NH₃. Therefore, 0.901 mol of H₂ will produce:
[
\text{Moles of NH₃} = \frac{2}{3} \times 0.901 \, \text{mol} = 0.601 \, \text{mol}
]
Step 4: Calculate the enthalpy change
The standard molar enthalpy of formation of NH₃(g) is -45.9 kJ/mol. The enthalpy change for the reaction is:
[
\Delta H = \text{moles of NH₃} \times \Delta H_f^\circ(\text{NH₃}) = 0.601 \, \text{mol} \times (-45.9 \, \text{kJ/mol}) = -27.6 \, \text{kJ}
]
Conclusion
The enthalpy change for the reaction is -27.6 kJ. This means that the reaction releases 27.6 kJ of energy when 9.10 g of N₂ and 1.82 g of H₂ react to form NH₃.