Consider the following reaction:
2H2O2(aq)?2H2O(l)+O2(g)
The graph (Figure 1) shows the concentration of H2O2 as a function of time. Use the graph to calculate the following.
A.)The average rate of the reaction between 10 and 20 s.
Express your answer using two significant figures.
B.) The instantaneous rate of the reaction at 30 s.
Express your answer using two significant figures.
C.)The instantaneous rate of formation of O2 at 50 s.
Express your answer using two significant figures.
D.)If the initial volume of the H2O2 solution is 1.6 L , what total amount of O2 (in moles) is formed in the first 50 s of reaction?
Express your answer using two significant figures.
The Correct Answer and Explanation is :
A) Average rate of reaction between 10 and 20 s
The average rate of the reaction over a period can be calculated using the following formula:
[
\text{Average rate} = \frac{\text{Change in concentration of H2O2}}{\text{Change in time}}
]
To do this, locate the concentration of H2O2 at 10 s and 20 s on the graph. Find the difference in concentration at these two points (Δ[H2O2]) and divide by the time interval (Δt = 20 s – 10 s).
Formula:
[
\text{Average rate} = \frac{[H2O2]{20s} – [H2O2]{10s}}{20s – 10s}
]
B) Instantaneous rate of the reaction at 30 s
To find the instantaneous rate at 30 s, you need to determine the slope of the tangent line to the concentration vs. time curve at this point. This can be approximated by using the data points near 30 s and calculating the slope between them.
The formula is similar:
[
\text{Instantaneous rate} \approx \frac{\Delta [H2O2]}{\Delta t}
]
You would look for a small time interval around 30 s (such as between 28 and 32 s) and calculate the slope of the curve in that range.
C) Instantaneous rate of formation of O2 at 50 s
The reaction indicates that 2 moles of H2O2 produce 1 mole of O2. Thus, the rate of O2 formation is half the rate of H2O2 consumption. To find the instantaneous rate of O2 formation, you can simply divide the instantaneous rate of H2O2 by 2:
[
\text{Rate of O2 formation} = \frac{1}{2} \times \text{Instantaneous rate of H2O2}
]
D) Total amount of O2 formed in the first 50 s
To find the total amount of O2 produced, you first need to determine the total change in the concentration of H2O2 over the first 50 seconds. Then, use the stoichiometry of the reaction to convert the change in H2O2 into the amount of O2 formed.
For example, if the concentration of H2O2 changes by Δ[H2O2] in 50 s, the amount of O2 formed will be:
[
\text{Moles of O2} = \frac{1}{2} \times \Delta[H2O2] \times \text{Volume (in L)}
]
You would multiply the concentration change by the volume (1.6 L) and divide by 2 (since 2 moles of H2O2 produce 1 mole of O2).
Now, I can also generate an image for the graph if you describe the concentration values at various times or provide more details. Let me know if you’d like me to help you with that as well!