Zero Order Reaction: Substrate Utilization Time

For a zero-order reaction, the relationship between concentration and time is linear.

The integrated rate law is given by:

$[A]_t = [A]_0 - kt$

Where:

• $[A]_t$ = Concentration at time $t$
• $[A]_0$ = Initial concentration
• $k$ = Rate constant
• $t$ = Time

Calculating the Rate Constant (k)

We are given:

• Initial concentration, $[A]_0 = 42 \text{ g L}^{-1}$
• Concentration after 4 hours, $[A]_t = 14 \text{ g L}^{-1}$
• Time, $t = 4 \text{ hours}$

Substitute these values into the integrated rate law:

$14 \text{ g L}^{-1} = 42 \text{ g L}^{-1} - k \times (4 \text{ hours})$

Rearrange to solve for $k$:

$k \times (4 \text{ hours}) = 42 \text{ g L}^{-1} - 14 \text{ g L}^{-1}$

$k \times (4 \text{ hours}) = 28 \text{ g L}^{-1}$

$k = \frac{28 \text{ g L}^{-1}}{4 \text{ hours}} = 7 \text{ g L}^{-1} \text{ hour}^{-1}$

Calculating Total Time for Utilization

We need to find the time ($t_{total}$) when the substrate is completely utilized, meaning the concentration $[A]_t$ becomes $0 \text{ g L}^{-1}$.

Using the integrated rate law with $[A]_t = 0$ and the calculated rate constant $k = 7 \text{ g L}^{-1} \text{ hour}^{-1}$:

$0 \text{ g L}^{-1} = 42 \text{ g L}^{-1} - (7 \text{ g L}^{-1} \text{ hour}^{-1}) \times t_{total}$

Rearrange to solve for $t_{total}$:

$(7 \text{ g L}^{-1} \text{ hour}^{-1}) \times t_{total} = 42 \text{ g L}^{-1}$

$t_{total} = \frac{42 \text{ g L}^{-1}}{7 \text{ g L}^{-1} \text{ hour}^{-1}}$

$t_{total} = 6 \text{ hours}$

The total time taken for complete utilization of the substrate is 6 hours.