For the reaction 2A B+C, Kc = 4 x 10^-3. At a given time, the composition of reaction mixture is : [A]=[B]=[C]=2 x 10^-3 M. Then, which of the following is correct?

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q55 – NEET-UG 2024

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Home Chemistry EquilibriumQuestion #436

For the reaction

2 A ⇌ B + C

K_{c} = 4 \times 10^{- 3}

. At a given time, the composition of reaction mixture is :

[A] = [B] = [C] = 2 \times 10^{- 3} M

. Then, which of the following is correct?

Clue 1 of 3: Reaction Quotient

To determine the direction a reaction will proceed, compare the reaction quotient ( $Q_{c}$ ) with the equilibrium constant ( $K_{c}$ ).

🧠 Full Solution Path

Step 1: Calculate the Reaction Quotient (

Q_{c}

)✦ Active

For the given reaction $2 A ⇌ B + C$ , the reaction quotient $Q_{c}$ is expressed as:

Q_{c} = \frac{[B] [C]}{[A]^{2}}

Substitute the given concentrations: $[A] = [B] = [C] = 2 \times 10^{- 3} M$

Q_{c} = \frac{(2 \times 10^{- 3}) (2 \times 10^{- 3})}{(2 \times 10^{- 3})^{2}} = \frac{4 \times 10^{- 6}}{4 \times 10^{- 6}} = 1

Step 2: Compare

Q_{c}

with

K_{c}

○ Expand

The calculated reaction quotient is $Q_{c} = 1$ . The given equilibrium constant is $K_{c} = 4 \times 10^{- 3} = 0.004$ .

Comparing these values, we find that $Q_{c} > K_{c}$ ( $1 > 0.004$ ).

Step 3: Determine the Direction of Reaction○ Expand

When $Q_{c} > K_{c}$ , the concentration of products is higher than at equilibrium, and the concentration of reactants is lower. To reach equilibrium, the reaction must proceed in the reverse (backward) direction to decrease the product concentrations and increase the reactant concentrations.

💡 Teacher's Secret Hint

Remember the rules: $Q_{c} < K_{c}$ (forward), $Q_{c} > K_{c}$ (backward), $Q_{c} = K_{c}$ (equilibrium).

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Chemistry Question 55 – NEET-UG 2024

Clue 1 of 3: Reaction Quotient

🧠 Full Solution Path