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

Hint:

Recall that for a first-order reaction, the rate constant links to the half-life through a relation involving ln 2. Try deriving it yourself from the integrated rate law by setting the concentration equal to half its initial value at the half-life point, instead of just plugging into a memorised formula.

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.

Answer 3

Concept:

• Instead of starting from the ready-made half-life formula, the same result can be rebuilt directly from the integrated first-order rate law, using the definition of half-life as the time at which the concentration drops to exactly half its starting value.

Step 1: Write the integrated first-order rate law.
For a first-order reaction, $\ln\dfrac{[A]_0}{[A]}=kt$, where $[A]_0$ is the starting concentration and $[A]$ is the concentration at time $t$.

Step 2: Apply the definition of half-life.
At $t=t_{1/2}$, the concentration has fallen to exactly half, so $[A]=\dfrac{[A]_0}{2}$. Substituting this in:
$\ln\dfrac{[A]_0}{[A]_0/2}=k\,t_{1/2}$
$\ln 2=k\,t_{1/2}$

Step 3: Substitute the known numbers.
$\ln 2=0.693$ and $t_{1/2}=693$ sec, so:
$0.693=k\times693$

Step 4: Solve for k.
$k=\dfrac{0.693}{693}=0.001\ \text{sec}^{-1}$

Final Answer: $k=0.001\ \text{sec}^{-1}$, option A