Question

Consider the first order reaction R → P. The fraction of molecules decomposed in the given first order reaction can be expressed as

• A. \(1 - e^{-k_1 t}\)
• B. \(1 + e^{-k_1 t}\)
• C. \(1 + e^{k_1 t}\)
• D. \(1 - e^{k_1 t}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a first-order reaction, the concentration of the reactant decreases exponentially over time. We need to find the mathematical expression representing the fraction of the initial amount that has already reacted (decomposed).
Step 2: Key Formula or Approach:
The integrated rate law for a first-order reaction is:
\([\text{R}]_t = [\text{R}]_0 e^{-kt}\)
where \([\text{R}]_0\) is the initial concentration, \([\text{R}]_t\) is the concentration remaining at time \(t\), and \(k\) is the rate constant (often denoted \(k_1\) for first-order).
Step 3: Detailed Explanation:
The amount of reactant that has decomposed is the difference between the initial concentration and the concentration remaining:
\(\text{Amount decomposed} = [\text{R}]_0 - [\text{R}]_t\).
The "fraction of molecules decomposed" is this decomposed amount divided by the original amount:
\(\text{Fraction decomposed} = \frac{[\text{R}]_0 - [\text{R}]_t}{[\text{R}]_0}\)
Substitute the integrated rate law \([\text{R}]_t = [\text{R}]_0 e^{-k_1t}\) into the expression:
\(\text{Fraction decomposed} = \frac{[\text{R}]_0 - [\text{R}]_0 e^{-k_1t}}{[\text{R}]_0}\)
Factor out \([\text{R}]_0\) in the numerator:
\(\text{Fraction decomposed} = \frac{[\text{R}]_0 (1 - e^{-k_1t})}{[\text{R}]_0}\)
Cancel out \([\text{R}]_0\):
\(\text{Fraction decomposed} = 1 - e^{-k_1t}\).
Step 4: Final Answer:
The correct expression is \(1 - e^{-k_1t}\).