Question

Rate constant \(k\) for the first order reaction is \( 2.54 \times 10^{-3} \text{ s}^{-1} \). Calculate the time required for three-fourth of the reactant to decompose. (Given: \( \log 4 = 0.60 \))

Hint:

For a first-order reaction, the time required for \( 75\% \) completion (\(t_{3/4}\)) is exactly twice the half-life (\(t_{1/2}\)). You can also solve this by finding \( t_{1/2} \) first and multiplying by 2!

Answer 1

Concept: For a first-order reaction, the time required for a specific fraction of the reactant to decompose can be found using the integrated rate equation: \[ t = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]} \right) \] When a reaction is "three-fourths complete," it means that \( 3/4 \) of the initial concentration has reacted, and therefore \( 1/4 \) of the initial concentration remains. Step 1: Setting up the initial and final concentrations.
Let the initial concentration be \( [R]_0 = 100 \) (or \(a\)). Since three-fourths (\(75\%\)) of the reactant decomposes, the amount reacted is \( 75 \). The remaining concentration \( [R] \) is: \[ [R] = 100 - 75 = 25 \quad \left(\text{or } [R] = a - \frac{3a}{4} = \frac{a}{4} \right) \]

Step 2: Substituting the values into the integrated rate equation.
Given rate constant \( k = 2.54 \times 10^{-3} \text{ s}^{-1} \). \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log \left( \frac{100}{25} \right) \]

Step 3: Simplifying the logarithmic term and calculating time.
\[ \frac{100}{25} = 4 \] \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log (4) \] We are given \( \log 4 = 0.60 \). Substitute this into the equation: \[ t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} \] \[ t = \frac{1.3818}{2.54 \times 10^{-3}} \] \[ t = 544.015 \text{ s} \] Rounding to the nearest whole number gives \( 544 \text{ s} \). Final Answer: The time required for three-fourth decomposition is \( 544 \text{ s} \).