Fraction of Molecules Decomposed in First Order Reaction

For a first order chemical reaction: $\text{R} \rightarrow \text{P}$

The rate law for a first order reaction is given by the differential rate equation: $-\frac{d[\text{R}]}{dt} = k_1 [\text{R}]$

where:

• $[\text{R}]$ is the concentration of the reactant $\text{R}$ at time $t$,
• $k_1$ is the first order rate constant.

Rearranging and integrating from time $t = 0$ (where $[\text{R}] = [\text{R}]_0$) to time $t$ (where $[\text{R}] = [\text{R}]_t$): $\int_{[\text{R}]_0}^{[\text{R}]_t} \frac{d[\text{R}]}{[\text{R}]} = -k_1 \int_0^t dt$

$\ln\left(\frac{[\text{R}]_t}{[\text{R}]_0}\right) = -k_1 t$

Taking the exponential on both sides yields the integrated rate law: $[\text{R}]_t = [\text{R}]_0 e^{-k_1 t}$

The fraction of reactant molecules remaining unreacted at time $t$ is: $\text{Fraction remaining} = \frac{[\text{R}]_t}{[\text{R}]_0} = e^{-k_1 t}$

Therefore, the fraction of molecules decomposed (or reacted) by time $t$ is given by: $\text{Fraction decomposed} = 1 - \text{Fraction remaining} = 1 - \frac{[\text{R}]_t}{[\text{R}]_0} = 1 - e^{-k_1 t}$

Hence, the correct option is D.