The standard molar enthalpy of formation of NH(g) is -45.9 kJ/mol.

The standard molar enthalpy of formation of NH
(g) is -45.9 kJ/mol. What is the enthalpy change if 6.31 g N
(s) and 1.96 g H
(g) react to produce NH
(g)?

The Correct Answer and Explanation is:

To determine the enthalpy change for the given reaction, we first need to write the reaction for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases: N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)N2​(g)+3H2​(g)→2NH3​(g)

The reaction involves the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). You are given the standard molar enthalpy of formation of ammonia (NH₃(g)) as -45.9 kJ/mol.

Step 1: Determine moles of nitrogen and hydrogen

You are given the following masses:

• 6.31 g of nitrogen (N₂)
• 1.96 g of hydrogen (H₂)

First, we calculate the moles of nitrogen and hydrogen:

• Molar mass of nitrogen (N₂) = 28.0 g/mol
• Molar mass of hydrogen (H₂) = 2.0 g/mol

For nitrogen: moles of N2=6.31 g28.0 g/mol=0.225 mol\text{moles of N}_2 = \frac{6.31 \, \text{g}}{28.0 \, \text{g/mol}} = 0.225 \, \text{mol}moles of N2​=28.0g/mol6.31g​=0.225mol

For hydrogen: moles of H2=1.96 g2.0 g/mol=0.98 mol\text{moles of H}_2 = \frac{1.96 \, \text{g}}{2.0 \, \text{g/mol}} = 0.98 \, \text{mol}moles of H2​=2.0g/mol1.96g​=0.98mol

Step 2: Determine limiting reactant

From the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to form 2 moles of NH₃. We now compare the mole ratio:

• For 0.225 moles of N₂, the required moles of H₂ would be:

0.225 mol N2×3=0.675 mol H20.225 \, \text{mol N}_2 \times 3 = 0.675 \, \text{mol H}_20.225mol N2​×3=0.675mol H2​

Since we have 0.98 moles of H₂ (which is more than 0.675 moles), nitrogen (N₂) is the limiting reactant.

Step 3: Calculate the enthalpy change

Since 1 mole of N₂ produces 2 moles of NH₃, 0.225 moles of N₂ would produce: 0.225 mol N2×2=0.45 mol NH30.225 \, \text{mol N}_2 \times 2 = 0.45 \, \text{mol NH}_30.225mol N2​×2=0.45mol NH3​

The standard enthalpy of formation for NH₃ is -45.9 kJ/mol. The total enthalpy change is: Enthalpy change=0.45 mol×(−45.9 kJ/mol)=−20.66 kJ\text{Enthalpy change} = 0.45 \, \text{mol} \times (-45.9 \, \text{kJ/mol}) = -20.66 \, \text{kJ}Enthalpy change=0.45mol×(−45.9kJ/mol)=−20.66kJ

Final Answer:

The enthalpy change for the reaction is -20.66 kJ.

This negative value indicates that the reaction is exothermic, meaning that heat is released when ammonia is formed from nitrogen and hydrogen.