The initial concentration of a reactant in a zero order is 2.5M. calculate the final concentration of the reactant after 80sec if the rate constants is 0.0034 m/s
To calculate the final concentration of the reactant after 80 seconds in a zero-order reaction, we need to use the zero-order rate equation:
Rate = k
Where:
Rate is the rate of reaction
k is the rate constant
In this case, the rate constant is given as 0.0034 m/s.
Since it is a zero-order reaction, the rate is constant and does not depend on the initial concentration of the reactant.
To find the final concentration of the reactant, we need to calculate the amount of reactant consumed over the given time period.
Amount of reactant consumed = Rate * Time
Substituting the given values:
Amount of reactant consumed = 0.0034 m/s * 80 s = 0.272 moles
The final concentration of the reactant can be calculated by subtracting the amount of reactant consumed from the initial concentration:
Final concentration = Initial concentration - Amount of reactant consumed
Substituting the given values:
Final concentration = 2.5 M - 0.272 mol / (80 s) = 2.5 M - 0.0034 M = 2.4966 M
Therefore, the final concentration of the reactant after 80 seconds in a zero-order reaction is approximately 2.4966 M.
To calculate the final concentration of a reactant after a certain time in a zero-order reaction, we need to use the zero-order rate equation:
\[ \text{Rate} = -\frac{d[A]}{dt} = k \]
Where \([A]\) represents the concentration of reactant A at a given time, \(t\). In a zero-order reaction, the rate is independent of the concentration of the reactant.
To find the final concentration of the reactant after a certain time, first, we need to determine the rate constant (\(k\)).
Given that the rate constant is \(0.0034 \, \text{m/s}\), we can use this value in the rate equation. Since the rate of the reaction is constant and independent of concentration, we can write:
\[ \text{Rate} = k = 0.0034 \, \text{m/s} \]
Now, we can integrate the rate equation to find the relationship between time and concentration. Integrating, we have:
\[ -\int_{[A]_0}^{[A]_t}\frac{d[A]}{[A]} = \int_0^t k \, dt \]
Simplifying the equation, we get:
\[ -\ln \frac{[A]_t}{[A]_0} = k \cdot t \]
Where \([A]_t\) is the concentration of the reactant at time \(t\) and \([A]_0\) is the initial concentration of the reactant.
Rearranging the equation to solve for \([A]_t\), we have:
\[ [A]_t = [A]_0 \cdot e^{-k \cdot t} \]
Given that the initial concentration \([A]_0\) is \(2.5 \, \text{M}\) and the time \(t\) is \(80 \, \text{s}\), we can substitute these values into the equation to find the final concentration \([A]_t\).
\[ [A]_t = 2.5 \, \text{M} \cdot e^{-(0.0034 \, \text{m/s}) \cdot (80 \, \text{s})} \]
Evaluating this expression, we can now calculate the final concentration of the reactant after 80 seconds in a zero-order reaction.