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For a certain first-order gas-hase reaction

$$\ce{A \rightarrow B + C}$$

the time required for half of an initial amount of A to decompose is 8.8 min. If the initial pressure of A is 400 mmHg, the time required to reduce the partial pressure of A to 50 mmHg is:

(a) 1.1 min
(b) 17.6 min
(c) 26.4 min
(d) 35.2 min
(e) 70.4 min

The correct answer is (c).

When I approached this problem, I started off thinking about it as a zeroth order reaction for simplicity.

$$\mathrm{400~mmHg \rightarrow 200~mmHg \rightarrow 100~mmHg \rightarrow 50~mmHg}$$

So, if this was a zeroth order reaction (not dependent on the amount of A remaining), then time would be $\mathrm{3 \cdot 8.8~min = 26.4~min}$. But, since this is a first order reaction, then [A] does factor into the half life and thus the time to reach $\mathrm{50~mmHg}$ should be slightly higher than $\mathrm{26.4~min}$.

However, this doesn't seem to be the case. Perhaps it's because we're using pressures instead of concentrations? But isn't pressure synonymous with concentration in the gas phase?

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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

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