Consider the reaction: 2 H₂O₂(aq) → 2 H₂O(l ) + O₂( g) The graph shows the concentration of H₂O₂ as a function of time.

Use the graph to calculate each quantity: c. the instantaneous rate of formation of O₂ at 50 s

Video transcript

Welcome back everyone. In this example, we need to estimate the instantaneous rate of formation of oxygen at 70 seconds for the reaction below based on our given graph. So we have two moles of di nitrogen pent oxide which produces four moles of nitrogen dioxide and one mole of oxygen. And according to our graph, we have time given in seconds on the X axis and the concentration of oxygen on our Y axis. And according to the prompt, the instantaneous rate of formation is at 70 seconds. So we wanna go ahead and refer to the following formula where we would find the rate of oxygen being formed, which is going to equal the change in formation of oxygen divided by the change in time that it takes the oxygen to form. And since we're using delta, we would say that we would take the final concentration of oxygen minus the initial concentration of oxygen and then divided by our final time T two minus the initial time being T one. So plugging in what we know for our rate, we would say that our final concentration of oxygen. So because we have to refer to the instantaneous rate of oxygen being at 70 seconds, we're going to look at the time that occurs after 70 which according to our graph is 80 seconds. So we would use 80 seconds as our final time. And that goes in our denominator. This will correspond to the final concentration of oxygen, which we can see is at this point of our Y axis. And as a percentage that's about 0.79% which we will convert to a decimal as 0.79. Now we would subtract from our initial concentration of oxygen which will occur because again, the prompt says the instantaneous rate is going to be for 70 seconds. We want to look at the time before 70 seconds, which on a graph is going to be at 60 seconds here and looking for the concentration of oxygen at 60 seconds, we would see that this corresponds to about point or 65% which we would convert to about 0.65. So we have about 0.65 concentration of oxygen at 60 seconds. So we need to just add that subtraction sign in the denominator and plugging this quo into our calculators. We would get that our rate is going to equal a value of seven times 10 to the negative third power. And recall that our units for concentration is molar and we still have per second in our denominator. And so this will actually complete this example as our final answer for our instantaneous rate of formation of oxygen at 70 seconds. So I hope that everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.