$2A \xrightarrow{k} B$ is a zero-order reaction, where $k = 1.0\text{ mol L}^{-1}\text{ min}^{-1}$. If the initial concentration of A is $2\text{ M}$, then the time taken to complete 75% of the reaction will be
Zero-Order Reaction Time Calculation
For a zero-order reaction, the reaction rate is constant and independent of reactant concentrations. The integrated rate law relates the concentration of a reactant at time $t$ to its initial concentration and the rate constant.
The integrated rate law for a zero-order reaction involving reactant A is:
$ [A]_t = [A]_0 - kt $
Where:
- $[A]_t$ is the concentration of reactant A at time $t$.
- $[A]_0$ is the initial concentration of reactant A.
- $k$ is the rate constant.
- $t$ is the time taken.
Calculating Remaining Reactant Concentration
The question asks for the time taken to complete 75% of the reaction. This means 75% of the initial reactant A has been consumed.
The concentration of A remaining, $[A]_t$, will be:
$ [A]_t = [A]_0 - 0.75 \times [A]_0 $
$ [A]_t = (1 - 0.75) \times [A]_0 $
$ [A]_t = 0.25 \times [A]_0 $
Given the initial concentration $[A]_0 = 2\text{ M}$:
$ [A]_t = 0.25 \times 2\text{ M} = 0.5\text{ M} $
Determining the Time for 75% Completion
We can now use the integrated rate law to solve for the time $t$. Rearrange the equation:
$ t = \frac{[A]_0 - [A]_t}{k} $
Substitute the given values:
- Initial concentration, $[A]_0 = 2\text{ M}$
- Concentration at time t, $[A]_t = 0.5\text{ M}$
- Rate constant, $k = 1.0\text{ mol L}^{-1}\text{ min}^{-1}$
Perform the calculation:
$ t = \frac{2\text{ M} - 0.5\text{ M}}{1.0\text{ mol L}^{-1}\text{ min}^{-1}} $
$ t = \frac{1.5\text{ M}}{1.0\text{ mol L}^{-1}\text{ min}^{-1}} $
$ t = 1.5\text{ min} $
Thus, the time taken for 75% completion of the zero-order reaction is 1.5 minutes.
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