A plot of 1/[A] versus time for the hypothetical reaction A →B+C yields a straight line. What is the order of the reaction? first-order second-order third-order O fourth-order b If it took 90. s for the concentration to drop 38 percent from its initial value of 0.50 M, what is the rate constant for the reaction? |L/(mol · s) Rate constant =

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### Reaction Order and Rate Constant Determination

#### (a) Reaction Order

A plot of \( \frac{1}{[\text{A}]} \) versus time for the hypothetical reaction:

\[ \text{A} \rightarrow \text{B} + \text{C} \]

yields a straight line. What is the order of the reaction?

- ○ first-order
- ● second-order ✔
- ○ third-order
- ○ fourth-order

The reaction is determined to be second-order because the plot of \( \frac{1}{[\text{A}]} \) versus time produces a straight line.

#### (b) Rate Constant Calculation

If it took 90.0 seconds for the concentration to drop 38 percent from its initial value of 0.50 M, what is the rate constant for the reaction?

Rate constant = \_\_\_\_ L/(mol·s)

Fill in the blank with the calculated rate constant value, using the information provided about the concentration change and time duration.
Transcribed Image Text:### Reaction Order and Rate Constant Determination #### (a) Reaction Order A plot of \( \frac{1}{[\text{A}]} \) versus time for the hypothetical reaction: \[ \text{A} \rightarrow \text{B} + \text{C} \] yields a straight line. What is the order of the reaction? - ○ first-order - ● second-order ✔ - ○ third-order - ○ fourth-order The reaction is determined to be second-order because the plot of \( \frac{1}{[\text{A}]} \) versus time produces a straight line. #### (b) Rate Constant Calculation If it took 90.0 seconds for the concentration to drop 38 percent from its initial value of 0.50 M, what is the rate constant for the reaction? Rate constant = \_\_\_\_ L/(mol·s) Fill in the blank with the calculated rate constant value, using the information provided about the concentration change and time duration.
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