The tabulated data were collected for this reaction at 500 °C: CH₃CN(g) → CH₃NC( g) b. What is the half-life for this reaction (at the initial concentration)?

Video transcript

Hey everyone in this example, we need to consider the hypothetical decomposition of a b forming a plus b. Were given the data below at 423.15 Kelvin and we need to calculate the half life of the reaction at the initial concentration. Before we can determine the half life of our reaction, we need to determine what order kinetics this reaction follows. And so we should recall our zero order integrated rate law, which we would recall is calculated by taking the concentration of our reactant in this case A B at a given time and setting that equal to negative one times our rate constant K. Multiplied by time, which is then added to the initial concentration of our react in A. B. And we want to go ahead and plot our concentration of are reacting A B at a given time versus time. Next we want to go ahead and recall. So this will be our first formula for our second formula. We want to recall our first order integrated rate law which we recall is found from taking the natural log of our concentration of are reacting A B at a given time, which we would set equal to negative. One times I rate constant K. Multiplied by time plus the natural log of our initial concentration of are reacting A B. And here we would plot the concentration of are reacting a B at a given time versus time. We would see on our calculators. That when we plot the data given on the left according to our first order integrated rate law that we have a linear graph. So this is a good sign, meaning that our reaction here is most likely following first order kinetics. However, we're going to continue on and recall our third integrated rate law, which we would recall as a second order integrated rate law. And we would recall that that is found by taking one over our initial concentration of a reactant at a given time and setting that equal to our rate constant K. Times time added to one over our initial concentration of our reactant A. B initially. And here we would plot one over our initial concentration of are reacting A B at a given time versus time. And we would see that we do not get a linear graph here from our second order nor from our zero order integrated rate law according to the data given. So we would say we don't have a linear graph. And then here we don't have a linear graph. However, because we stated that we have a linear graph from plotting our concentration of a B at a given time versus time, according to first order kinetics. We can say therefore our reaction is first order. And so we can recall that to calculate the half life for a first order reaction. We're going to take the half life and set that equal to the natural log of two divided by the rate constant K. So we're going to relate our first order integrated rate law to y equal to M X plus B. Where we would recognize that our slope m is going to equal negative one times our rate constant K. And so we can say that our rate constant K. Is equal to negative one times our slope. Where we would calculate our rate constant K. By taking negative one tends our slope given in the prompt or sorry, given in our calculators. According to our first order integrated rate law calculation equal to a value of negative 0.875 inverse seconds. Which because we're multiplying this by negative one, we would have now a positive value of 0.875 inverse seconds. So this comes from our calculator when we look up the value for our slope according to our linear graph. And so just so it's extra clear we would see that our equation of our line in our calculators according to our linear graph gives us why? Equal to negative 0.875. Which as we stated below, is our slope times X plus our y intercept 0.916 to nine. So now that we've calculated our value for our rate constant K. We can go ahead and recall that to calculate half life according to first order kinetics. We said we would take the natural log of two divided by our rate constant K. And so this means we would take the natural log of two divided by our rate constant, which above we stated at 0.875 inverse seconds. And so this would give us our half life equal to a value of 79.2 seconds. And so this would be our final answer as our half life at the initial concentration for our reaction of A be producing A plus B. So I hope that everything I explained was clear what's highlighted in yellow represents our final answer. If you have any questions, just leave them down below and I will see everyone in the next practice video.