Question

The half-life for a zero order reaction having $0.02\, M$ initial concentration of reactant is $100\, s$. The rate constant (in $mol \,L^{-1} s^{-1})$ for the reaction is

• A. $1.0 \times 10^{-4}$
• B. $2.0 \times 10^{-4}$
• C. $2.0 \times 10^{-3}$
• D. $1.0\times 10^{-2}$

Answer 1

The Correct Option is A

To find the rate constant for a zero-order reaction given its half-life, we can use the formula specific to zero-order kinetics:

\(t_{\frac{1}{2}} = \frac{[A]_0}{2k}\)

where \(t_{\frac{1}{2}}\) is the half-life, \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Given:

• Initial concentration, \([A]_0 = 0.02 \, M\)
• Half-life, \(t_{\frac{1}{2}} = 100 \, s\)

We need to find the rate constant \(k\). Rearranging the formula to solve for \(k\), we get:

\(k = \frac{[A]_0}{2t_{\frac{1}{2}}}\)

Substitute the given values into this equation:

\(k = \frac{0.02}{2 \times 100} = \frac{0.02}{200}\)

Calculating gives:

\(k = 0.0001 \, \text{mol} \, L^{-1} \, s^{-1}\)

Converting to scientific notation, we have: \(k = 1.0 \times 10^{-4} \, \text{mol} \, L^{-1} \, s^{-1}\)

Thus, the correct option is $1.0 \times 10^{-4}$.

Therefore, the correct rate constant considering the given options and the calculations is accurately determined.