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Kinetics and Half Lives Using Pressure
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Marco Pagliero
Kinetics and Half Lives Using Pressure
Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.
What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.
For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is
So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:
$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$
where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.
As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.
Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.
What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.
For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is
So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:
$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$
where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.
As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.
Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.
What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.
For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is
$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$
and the usual expression for half-life (where I've rearranged to solve for $k$) is
$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$
So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:
$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$
where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.
As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.
Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.
What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.
For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is
$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$
and the usual expression for half-life (where I've rearranged to solve for $k$) is
$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$
So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:
$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$
where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.
As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.
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Your approach was perfect! Infact thanks for reminding me this method. But your conception is wrong regarding order of reaction.
For your kind information,
Zero order reaction has half life directly proportional to its initial concentration
If [A]° (initial concentration) , K (rate constant)
Then
T (1/2)= [A]° / 2K
But
For first order reaction,
T (1/2)= 0.693 / K
And you can get now that in first order reaction the half life is totally independent of the initial concentration.
So the shortcut you applied for Zero Order reaction was actually valid for the First Order reaction.
Now I request you to go through the textbook once again thoroughly and get the right idea about Rates and Order of reactions.
Cheers
Your approach was perfect! Infact thanks for reminding me this method. But your conception is wrong regarding order of reaction.
For your kind information,
Zero order reaction has half life directly proportional to its initial concentration
If [A]° (initial concentration) , K (rate constant)
Then
T (1/2)= [A]° / 2K
But
For first order reaction,
T (1/2)= 0.693 / K
And you can get now that in first order reaction the half life is totally independent of the initial concentration.
So the shortcut you applied for Zero Order reaction was actually valid for the First Order reaction.
Now I request you to go through the textbook once again thoroughly and get the right idea about Rates and Order of reactions.
Cheers
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