Question

The concentration of the reactant is reduced from \( 0.6 \text{ mol L}^{-1} \) to \( 0.2 \text{ mol L}^{-1} \) in \( 5 \text{ minutes} \) in a first order reaction. Calculate rate constant of the reaction. (Given: \( \log 3 = 0.48 \))

Hint:

When units of time are given in minutes, the rate constant for a first-order reaction will have units of \(\text{min}^{-1}\). You don't need to convert to seconds unless explicitly asked!

Answer 1

Concept: For a first-order reaction, the rate of reaction is directly proportional to the first power of the concentration of the reactant. The integrated rate equation that relates the rate constant (\(k\)), time (\(t\)), initial concentration (\([R]_0\)), and final concentration (\([R]\)) is given by: \[ k = \frac{2.303}{t} \log \left( \frac{[R]_0}{[R]} \right) \] This formula allows us to calculate the rate constant if we know how much reactant is left after a certain period. Step 1: Extracting the given values from the problem.

• Initial concentration of reactant, \( [R]_0 = 0.6 \text{ mol L}^{-1} \)

• Final concentration of reactant, \( [R] = 0.2 \text{ mol L}^{-1} \)

• Time taken, \( t = 5 \text{ minutes} \)

Step 2: Substituting the values into the integrated rate equation.
\[ k = \frac{2.303}{5 \text{ min}} \log \left( \frac{0.6}{0.2} \right) \]

Step 3: Simplifying the logarithmic term and calculating the result.
\[ \frac{0.6}{0.2} = 3 \] So the equation becomes: \[ k = \frac{2.303}{5} \log (3) \] We are given \( \log 3 = 0.48 \). Substituting this value: \[ k = \frac{2.303 \times 0.48}{5} \] \[ k = \frac{1.10544}{5} \] \[ k = 0.221088 \text{ min}^{-1} \] Rounding off to three significant figures, we get \( 0.221 \text{ min}^{-1} \). Final Answer: The rate constant of the reaction is \( 0.221 \text{ min}^{-1} \).