```
18) For the second-order reaction $A \rightarrow$ products, the following data are obtained:
$\begin{array}{l}
{[\mathrm{A}]=3.024 \mathrm{M}, t=0 \mathrm{~min}} \\
{[\mathrm{~A}]=2.935 \mathrm{M}, t=1.0 \mathrm{~min}} \\
{[\mathrm{~A}]=2.852 \mathrm{M}, t=2.0 \mathrm{~min}}
\end{array}$
What is the initial rate of the reaction?
A) $0.089 \mathrm{M} \mathrm{min}-1$
B) $0.030 \mathrm{M} \mathrm{min}-1$
C) $11.2 \mathrm{M}$ min -1
D) $0.010 \mathrm{M} \mathrm{min}-1$
E) $0.091 \mathrm{M} \mathrm{min}^{-1}$
```

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Answer from Sia

Posted 4 months ago

D

Key Concept

Rate Law for Second-Order Reactions

Explanation

The rate of a second-order reaction can be determined using the rate law, which for a second-order reaction with respect to A is given by the equation $rate = k[A]^2$. To find the initial rate, we use the initial concentration of A and calculate the change in concentration over the change in time, which is the slope of the line when plotting $1/[A]$ versus time. The initial rate is the rate at $t = 0$, which can be approximated by the change in concentration between $t = 0$ and $t = 1.0$ minute. Using the given concentrations, the initial rate is approximately $0.089 \, \text{M} \cdot \text{min}^{-1}$ divided by the square of the initial concentration, which is $(3.024 \, \text{M})^2$. This calculation yields an initial rate of approximately $0.010 \, \text{M} \cdot \text{min}^{-1}$.

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