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MCQEasyJEE 2024Equilibrium Basics

JEE Chemistry 2024 Question with Solution

The equilibrium constant for the formation of NH₃ from N₂ and H₂, given [N₂] = 2×10−2 M2\times10^{-2} \, \text{M}, [H₂] = 3×10−2 M3\times10^{-2} \, \text{M}, [NH₃] = 1.5×10−2 M1.5\times10^{-2} \, \text{M} at 500 K500 \, \text{K}, is:

  • A

    417417

  • B

    41704170

  • C

    41.741.7

  • D

    4.174.17

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The equilibrium reaction is

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})

with [N2\text{N}_2] = 2×10−2 M2 \times 10^{-2} \, \text{M}, [H2\text{H}_2] = 3×10−2 M3 \times 10^{-2} \, \text{M} and [NH3\text{NH}_3] = 1.5×10−2 M1.5 \times 10^{-2} \, \text{M}.

Find: The value of KcK_c.

For the reaction,

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Substituting the given concentrations,

Kc=(1.5×10−2)2(2×10−2)(3×10−2)3K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}

Now calculate the numerator,

(1.5×10−2)2=2.25×10−4(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}

Next, calculate the denominator,

(2×10−2)(3×10−2)3=2×10−2×27×10−6=54×10−8=5.4×10−7(2 \times 10^{-2})(3 \times 10^{-2})^3 = 2 \times 10^{-2} \times 27 \times 10^{-6} = 54 \times 10^{-8} = 5.4 \times 10^{-7}

Therefore,

Kc=2.25×10−45.4×10−7≈416.67K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}} \approx 416.67

So, Kc≈417K_c \approx 417. Therefore, the correct option is A.

Direct Substitution

Given: [N2\text{N}_2] = 2×10−2 M2 \times 10^{-2} \, \text{M}, [H2\text{H}_2] = 3×10−2 M3 \times 10^{-2} \, \text{M}, [NH3\text{NH}_3] = 1.5×10−2 M1.5 \times 10^{-2} \, \text{M}.

Find: The equilibrium constant KcK_c.

Use the expression directly:

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Substitute the values,

Kc=(1.5×10−2)2(2×10−2)(3×10−2)3=417K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3} = 417

Therefore, the correct option is A.

Common mistakes

  • Using the wrong equilibrium expression is a common mistake. The powers of concentration must come from the balanced equation, so [H2\text{H}_2] must be raised to 33 and [NH3\text{NH}_3] to 22. Always write the balanced reaction first.

  • Students often forget to square 1.5×10−21.5 \times 10^{-2} correctly. Both the coefficient and the power of 1010 must be squared, giving 2.25×10−42.25 \times 10^{-4}, not 1.5×10−41.5 \times 10^{-4}.

  • Another mistake is handling scientific notation incorrectly in the denominator. Since (3×10−2)3=27×10−6(3 \times 10^{-2})^3 = 27 \times 10^{-6}, multiplying by 2×10−22 \times 10^{-2} gives 54×10−8=5.4×10−754 \times 10^{-8} = 5.4 \times 10^{-7}. Track powers of 1010 carefully.

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