Question

A first-order reaction has a half-life of 693 sec. What will be its rate constant?

• A. 0.001 sec^-1
• B. 1 sec^-1
• C. 0.01 sec^-1
• D. 0.1 sec^-1

Answer 1

The Correct Option is A

The rate constant for a first-order reaction can be calculated using the formula for half-life: t_1/2 = 0.693/k.

Given: \(t_{\frac {1}{2}} = 693\) sec.

Substituting in the half-life formula: \(693 = \frac {0.693}{k}\).

Rearrange to find k:\( k = \frac {0.693}{693}\).

Simplifying:\( k = 0.001\ sec^{-1}\).

Therefore, the rate constant is \(0.001\ sec^{-1}\).

Answer 2

For a first-order reaction, the relationship between the rate constant kkk and the half-life (t1/2t_{1/2}t1/2​) is given by the formula:

\(t_{1/2} = \frac{0.693}{k}\)

Where:

\(t_{1/2}\) is the half-life of the reaction.

\(k\) is the rate constant.

Given:

\(t_{1/2} = 693\) sec

We can solve for the rate constant \(k\):

\({t_{1/2}} = \frac{0.693}{693} = 0.001\ sec{−1}\)

Thus, the rate constant is 0.001 sec⁻¹, which corresponds to Option A.