Question

1. during a chemical reaction, the concentration of reactant a changes over time. the initial concentration of a is 0.6 mol/l. after 20 seconds, it decreases to 0.3 mol/l. calculate the instantaneous rate of reaction for reactant a during this time interval.

a. 0.015 mol/ls
b. 0.03 mol/ls
c. 0.010 mol/ls
d. 0.015 mol/ls
e. 0.0015 mol/ls

Explanation:

Step1: Recall the formula for average rate of reaction

The average rate of reaction for a reactant is given by the change in concentration divided by the change in time. The formula is \( \text{Rate} = \frac{\Delta [A]}{\Delta t} \), where \( \Delta [A] \) is the change in concentration of A and \( \Delta t \) is the change in time.

Step2: Calculate the change in concentration

The initial concentration \( [A]_i = 0.6 \, \text{mol/L} \) and the final concentration \( [A]_f = 0.3 \, \text{mol/L} \). So, \( \Delta [A] = [A]_f - [A]_i = 0.3 - 0.6 = -0.3 \, \text{mol/L} \). The negative sign indicates a decrease in concentration.

Step3: Calculate the change in time

The time interval \( \Delta t = 20 \, \text{s} \).

Step4: Calculate the rate of reaction

Using the formula \( \text{Rate} = \frac{\Delta [A]}{\Delta t} \), we substitute the values: \( \text{Rate} = \frac{-0.3}{20} = -0.015 \, \text{mol/L·s} \). The negative sign indicates the consumption of the reactant, but the question asks for the rate (magnitude or considering the decrease), so the rate is \( 0.015 \, \text{mol/L·s} \) (or \( 0.015 \, \text{mol/L·s} \), which is option a? Wait, wait, let's check the options again. Wait, the options are in mol/L·s? Wait, the options are a. 0.015 mol/L·s, b. 0.030 mol/L·s, c. 0.010 mol/L·s, d. 0.015 mol/L·s? Wait, maybe a typo, but let's recalculate. Wait, \( \Delta [A] = 0.6 - 0.3 = 0.3 \, \text{mol/L} \) (taking the absolute value for rate of decrease), then \( \text{Rate} = \frac{0.3}{20} = 0.015 \, \text{mol/L·s} \). So the correct answer is a (or the option with 0.015 mol/L·s).

Answer:

a. 0.015 mol/L·s