Question

For a first-order reaction, if the rate constant is \( k = 6.93 \times 10^{-3} \), what is the half-life (\( t_{1/2} \)) of the reaction?

• A. \( t_{1/2} = \frac{0.693}{k} \)
• B. \( t_{1/2} = \frac{1}{k} \)
• C. \( t_{1/2} = \frac{2}{k} \)
• D. \( t_{1/2} = k \)

Hint:

For first-order reactions, remember that the half-life is inversely proportional to the rate constant \( k \). The larger the value of \( k \), the shorter the half-life.

Answer 1

The Correct Option is A

For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant of the reaction. Given that \( k = 6.93 \times 10^{-3} \), we can substitute this value into the equation: \[ t_{1/2} = \frac{0.693}{6.93 \times 10^{-3}} = 100 \text{ seconds} \] Thus, the correct formula for the half-life of a first-order reaction is \( t_{1/2} = \frac{0.693}{k} \).