Question

A general reaction written as 2b - c 2d is studied and yields the following?

Answer 1

The rate law for the given reaction A + 2B → C + 2D is
`\( \text{rate} = k[A][B]^2 \)`, where
`\( k \)` is the rate constant. The initial rate of the reaction
`(\( \text{Initial rate} = \Delta[C]/\Delta t \)) is 8.00 x 10\(^(-3)\) mol/Ls`, with initial concentrations of
`\( [A]_0 = 0.150 \, \text{M} \) and \( [B]_0 = 0.150 \, \text{M} \).`

Step-by-step explanation

In the study of reaction kinetics, the rate law expresses the relationship between the rate of a chemical reaction and the concentrations of its reactants. For the given reaction
`\( A + 2B \rightarrow C + 2D \)`, the rate law is
`\( \text{rate} = k[A][B]^2 \)`, indicating that the reaction is second order with respect to
`\( B \)` and first order with respect to
`\( A \)`. The given initial rate
`(\( \Delta[C]/\Delta t \))` and initial concentrations
`(\( [A]_0 = 0.150 \, \text{M} \)` and
`\( [B]_0 = 0.150 \, \text{M} \))` allow us to determine the rate constant
`(\( k \))`.

Using the formula
`\( \text{rate} = k[A][B]^2 \)`, we can rearrange it to solve for
`\( k \)`:

`\[ k = \frac{\text{rate}}{[A][B]^2} \]`

Substituting the given values:

`\[ k = \frac{8.00 * 10^(-3) \, \text{mol/Ls}}{(0.150 \, \text{M})(0.150 \, \text{M})^2} \]`

Calculating this yields the rate constant
`\( k \)`. The knowledge of
`\( k \)` allows for predicting the rate of the reaction under different conditions and gaining insights into the reaction mechanism. Understanding the rate law is crucial for optimizing reaction conditions in practical applications, such as in industrial processes or pharmaceutical development.

Complete Question here:

A general reaction written as A + 2B C+2D is studied and yields the following data: B]o Initial Δ[C]/At [A]0 0.150 M 0.150 M 8.00 x 10-3 mol/Ls

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