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 Information:

• 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
• Equilibrium Moles of \(\mathrm{NH_3}\) formed: 0.4 moles

Step 1: Determine Moles Change
Let \( x \) represent the moles of \(\mathrm{N_2}\) reacted. Stoichiometry dictates the changes:

	\(\mathrm{N_2}\)	\(\mathrm{H_2}\)	\(\mathrm{NH_3}\)
Initial (moles)	1	3	0
Change (moles)	-x	-3x	+2x
Equilibrium (moles)	1-x	3-3x	2x

 

Step 2: Calculate \( x \)
At equilibrium, \(\mathrm{NH_3}\) is 0.4 moles:

\[ 2x = 0.4 \implies x = 0.2 \text{ moles} \]

 

Step 3: Calculate Equilibrium Moles of \(\mathrm{H_2}\)
Substitute \( x = 0.2 \) into the \(\mathrm{H_2}\) equilibrium expression:

\[ \text{Moles of } \mathrm{H_2} = 3 - 3x = 3 - 3(0.2) = 3 - 0.6 = 2.4 \text{ moles} \]

 

Step 4: Verification of Equilibrium Moles

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

Conclusion
The equilibrium concentration of \(\mathrm{H_2}\) is \(\boxed{2.4 \text{ moles}}\).

Final Answer
The calculated equilibrium concentration of \(\mathrm{H_2}\) is:

\[ \boxed{2.4 \text{ moles}} \]