Question

For a reaction A + 2B → Products, when concentration of B alone is increased half life remains the same. If concentration of A alone is doubled, rate remains the same. The unit of rate constant for the reaction is

• A. S^-1
• B. L mol^-1 S^-1
• C. mol L^-1 S^-1
• D. atm^-1

Answer 1

The Correct Option is A

1. Determine the rate law based on the given information
The general rate law for the reaction A + 2B → Products can be written as:

Rate = k[A]^m[B]ⁿ

where:
- k is the rate constant
- m is the order with respect to A
- n is the order with respect to B

We are given two pieces of information:

1. When the concentration of B alone is increased, the half-life remains the same. This indicates that the reaction is first order with respect to B. Thus, n = 1.
2. If the concentration of A alone is doubled, the rate remains the same. This indicates that the reaction is zero order with respect to A. Thus, m = 0.

Therefore, the rate law is:

Rate = k[A]⁰[B]¹ = k[B]

2. Determine the units of the rate constant
Since Rate = k[B], we can write k = Rate/[B]

- The units of rate are typically mol L^-1 s^-1
- The units of concentration ([B]) are mol L^-1

Therefore, the units of k are:

k = (mol L^-1 s^-1) / (mol L^-1) = s^-1

Final Answer:
(A) s^-1

Answer 2

The reaction is of the form: \(A + 2B → Products\).

When the concentration of B is increased, and the half-life remains the same, it indicates that the reaction is zero-order with respect to B. This is because in a zero-order reaction, the rate is independent of the concentration of B.

When the concentration of A is doubled and the rate remains the same, it suggests that the reaction is also zero-order with respect to A, because in a zero-order reaction, the rate is independent of the concentration of the reactant.

For a zero-order reaction, the unit of the rate constant (k) is \(S^{-1}\).

The correct answer is (A) : \(S^{-1}.\)