Chemistry Q131 – AP-EAMCET 2025

T (K)

, the equilibrium constant for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)

6 \times 10^{- 2}

. At equilibrium if the molar concentrations of

H_{2}

and

{NH}_{3}

are

0.25 M

and

0.06 M

respectively, the equilibrium concentration of dinitrogen (in

{mol L}^{- 1}

) is

Clue 1 of 3: Equilibrium Constant Expression

The equilibrium constant ( $K$ ) for a reversible reaction relates the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients.

🧠 Full Solution Path

Step 1: Write the Equilibrium Constant Expression✦ Active

The given reversible reaction is:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)

The equilibrium constant expression ( $K$ ) for this reaction is defined as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients:

K = \frac{[{NH}_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}

💡 Teacher's Secret Hint

Remember that only gaseous and aqueous species are included in the equilibrium constant expression. Pure solids and liquids are omitted.

Step 2: Rearrange to Solve for Dinitrogen Concentration○ Expand

We need to find the equilibrium concentration of dinitrogen, $[N_{2}]$ . We can rearrange the equilibrium constant expression to solve for $[N_{2}]$ :

[N_{2}] = \frac{[{NH}_{3}]^{2}}{K [H_{2}]^{3}}

Step 3: Substitute Given Values○ Expand

Now, substitute the given values into the rearranged equation:

Equilibrium constant, $K = 6 \times 10^{- 2}$

Equilibrium concentration of ammonia, $[{NH}_{3}] = 0.06 M$

Equilibrium concentration of hydrogen, $[H_{2}] = 0.25 M$

[N_{2}] = \frac{(0.06)^{2}}{(6 \times 10^{- 2}) (0.25)^{3}}

💡 Teacher's Secret Hint

Pay close attention to the powers (stoichiometric coefficients) when substituting the concentrations.

Step 4: Calculate the Dinitrogen Concentration○ Expand

Perform the calculation:

[N_{2}] = \frac{0.0036}{(0.06) (0.015625)}

[N_{2}] = \frac{0.0036}{0.0009375}

[N_{2}] = 3.84 {mol L}^{- 1}

The equilibrium concentration of dinitrogen is $3.84 {mol L}^{- 1}$ .

💡 Teacher's Secret Hint

Double-check your calculations, especially with exponents and scientific notation, to avoid arithmetic errors.

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Chemistry Question 131 – AP-EAMCET 2025

Clue 1 of 3: Equilibrium Constant Expression

🧠 Full Solution Path