JEE Main 2024 · 8 April · evening · Single correct · +4 / −1
For a reaction A →(K₁) B →(K₂) C If the rate of formation of B is set to be zero then the concentration of B is given by :
- (a) (K₁-K₂)[A]
- (b) (K₁+K₂)[A]
- (c) K₁ K₂[ A]
- (d) (K₁ / K₂)[A]
Answer and solution
Answer (d) (K₁ / K₂)[A]
Solution
Chemistry ma’am · 3 steps
Rate of formation of B is zero means B is made exactly as fast as it is used up.
- Two terms for B. B is formed from A at rate k₁[A] and consumed to give C at rate k₂[B], both first order. The net rate for B is the difference of the two.
d[B]/dt = k₁[A] - k₂[B] = 0
- Solve for B. Setting the net rate to zero gives k₁[A] = k₂[B], so [B] = k₁/k₂[A]. This is the steady-state approximation for the intermediate B.
- So. The concentration of B is (k₁/k₂)[A], which is option (d).
The idea to keep. For an intermediate, write rate of formation minus rate of removal, set it to zero and solve. Each step contributes its own rate constant times the concentration of its own reactant.
Answer from the NTA final answer key. Solution by PrepWiser.