Question

here is a graph of the pressure of oxygen (o₂) in a reaction vessel during a certain chemical reaction. use this graph to answer the questions in the table below. is o₂ being created or destroyed by the chemical reaction? created destroyed neither created nor destroyed if o₂ is being created or destroyed, what is the rate at which it is being created or destroyed 60 seconds after the reaction starts? round your answer to 2 significant digits. also be sure your answer has the correct unit symbol. if o₂ is being created or destroyed, what is the average rate at which it is being created or destroyed during the first 60 seconds of the reaction? round your answer to 2 significant digits. also be sure your answer has the correct unit symbol.

Explanation:

Step1: Determine if O₂ is created or destroyed

Observe the pressure - time graph. The pressure of O₂ decreases over time. So, O₂ is being destroyed.

Step2: Calculate the instantaneous rate at 60 s

To find the instantaneous rate at 60 s, we need to find the slope of the tangent line to the curve at \(t = 60\ s\).
We can estimate the slope by taking two points very close to \(t=60\ s\). Let's take two points: \((t_1,p_1)=(50,1.75)\) and \((t_2,p_2)=(70,1.25)\) (approximate values from the graph). The slope \(m=\frac{\Delta p}{\Delta t}=\frac{p_2 - p_1}{t_2 - t_1}=\frac{1.25 - 1.75}{70 - 50}=\frac{- 0.5}{20}=-0.025\ atm/s\).

Step3: Calculate the average rate in the first 60 s

The initial pressure \(p_0\) at \(t = 0\ s\) is approximately \(2.75\ atm\), and the pressure \(p\) at \(t = 60\ s\) is approximately \(1.75\ atm\). The average rate \(r=\frac{\Delta p}{\Delta t}=\frac{p - p_0}{t - 0}=\frac{1.75 - 2.75}{60}=\frac{-1}{60}\approx - 0.017\ atm/s\).

Answer:

1. O₂ is being destroyed.
2. - 0.025 atm/s
3. - 0.017 atm/s