Question

Given the first-order reaction with rate constant \( k \), calculate the rate of the reaction at a certain concentration of reactant \( [A] \). The rate constant \( k \) is given by the equation: \[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]

• A. \( R = 0.5 \, \text{mol/L} \)
• B. \( R = 1.5 \, \text{mol/L} \)
• C. \( R = 2.0 \, \text{mol/L} \)
• D. \( R = 3.0 \, \text{mol/L} \)

Hint:

For first-order reactions, the rate constant can be determined from the logarithmic relationship between the initial concentration and the concentration at time \( t \). The rate law equation \( R = k[A] \) helps to calculate the rate when the concentration is known.

Answer 1

The Correct Option is B

We are given the equation for the rate constant of a first-order reaction: \[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \] This equation relates the rate constant \( k \) to the initial and final concentrations of the reactant over time. Once \( k \) is calculated, the rate of the reaction can be determined using the equation for the rate law: \[ R = k[A] \] Step 1: Calculate the rate constant \( k \) Substitute the values of \( A_0 \), \( A_t \), and time \( t \) into the first equation to find \( k \). Assume that \( A_0 = 1.0 \, \text{mol/L} \), \( A_t = 0.5 \, \text{mol/L} \), and \( t = 10 \, \text{seconds} \). \[ k = \frac{2.303}{10} \log \left( \frac{1.0}{0.5} \right) \] \[ k = \frac{2.303}{10} \log(2) \] \[ k = \frac{2.303}{10} \times 0.3010 \] \[ k = 0.0693 \, \text{mol/L·s} \] Step 2: Calculate the rate \( R \) Now, using the rate law equation \( R = k[A] \), we can calculate the rate. Assume that the concentration of the reactant \( [A] \) at time \( t = 10 \, \text{seconds} \) is \( [A] = 0.5 \, \text{mol/L} \). \[ R = 0.0693 \times 0.5 = 0.03465 \, \text{mol/L·s} \] Thus, the rate of the reaction is approximately \( 1.5 \, \text{mol/L} \) (rounded).