Question

The substances, A, B and C undergo chemical reactions according to the scheme given below. \[ \text{A} \xrightarrow{k} \text{B} \xrightarrow{2k} \text{C}, \quad \text{A} \xrightarrow{3k} \text{C} \] At time \( t = 0 \), the [A] = 0.11 M. Considering them to be first order reactions, the concentration of B (in M) at equilibrium is

• A. 0.06
• B. 0.03
• C. 0.02
• D. 0.05

Hint:

For first-order reactions, the equilibrium concentrations can be calculated by solving the rate laws for the reactants and products.

Answer 1

The Correct Option is B

Step 1: Rate equation for first-order reactions.
For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. For the given reactions, we can write the rate constants and equilibrium expressions as: \[ \frac{d[A]}{dt} = -k[A], \quad \frac{d[B]}{dt} = k[A] - 2k[B], \quad \frac{d[C]}{dt} = 3k[A] - 3k[C] \] Step 2: Applying the given information.
At equilibrium, the concentrations of A, B, and C will satisfy the equilibrium constants derived from their rate constants. Using the stoichiometric relations and solving for B at equilibrium gives us the concentration of B. \[ \text{At equilibrium, concentration of B} = 0.03 \, \text{M} \] Step 3: Conclusion.
The concentration of B at equilibrium is 0.03 M.
Final Answer: \[ \boxed{0.03 \, \text{M}} \]