Question

For the reaction $ N_2 + 3H_2 \rightleftharpoons 2NH_3 $, if initially 1 mole of $ N_2 $ and 3 moles of $ H_2 $ are taken and at equilibrium 0.4 moles of $ NH_3 $ are formed, find the equilibrium concentration of $ H_2 $.

• A. 2.4 moles
• B. 2.8 moles
• C. 3.4 moles
• D. 3.0 moles

Hint:

Always use mole ratios from the balanced equation to compute changes in moles at equilibrium.

Answer 1

The Correct Option is A

Given:
 

• Chemical equation: \(\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}\)
• Initial moles:
  • \(\mathrm{N_2}\): 1 mole
  • \(\mathrm{H_2}\): 3 moles
  • \(\mathrm{NH_3}\): 0 moles (initially)
• At equilibrium: 0.4 moles of \(\mathrm{NH_3}\) formed

Step 1: Establish the Change in Moles
Let \( x \) be the moles of \(\mathrm{N_2}\) that react. According to the stoichiometry: \[ \begin{array}{lcc} & \mathrm{N_2} & + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \\ \text{Initial (moles):} & 1 & 3 \quad 0 \\ \text{Change (moles):} & -x & -3x \quad +2x \\ \text{Equilibrium (moles):} & 1-x & 3-3x \quad 2x \\ \end{array} \] 
Step 2: Solve for \( x \)
Given that at equilibrium, 0.4 moles of \(\mathrm{NH_3}\) are present: \[ 2x = 0.4 \implies x = 0.2 \text{ moles} \] 
Step 3: Calculate Equilibrium Moles of \(\mathrm{H_2}\)
Substitute \( x = 0.2 \) into the equilibrium expression for \(\mathrm{H_2}\): \[ \text{Moles of } \mathrm{H_2} = 3 - 3x = 3 - 3(0.2) = 3 - 0.6 = 2.4 \text{ moles} \] 
Step 4: Verification

• \(\mathrm{N_2}\) at equilibrium: \( 1 - 0.2 = 0.8 \) moles
• \(\mathrm{H_2}\) at equilibrium: \( 3 - 0.6 = 2.4 \) moles
• \(\mathrm{NH_3}\) at equilibrium: \( 2 \times 0.2 = 0.4 \) moles

Conclusion
The equilibrium concentration of \(\mathrm{H_2}\) is \(\boxed{2.4 \text{ moles}}\). 
Final Answer
Based on standard equilibrium calculations, the correct equilibrium concentration of \(\mathrm{H_2}\) is: \[ \boxed{2.4 \text{ moles}} \]