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From my GenChem practice exam:

Consider the following reaction:
$\ce{C4H8 (g) → 2C2H4 (g)}$
The first-order decomposition of cyclobutane to two molecules of ethene has a rate constant of $\ce{9.20\times10^{-3} s^{-1}}$. When an initial $\ce{2.26}$ moles of cyclobutane were allowed to decompose in a $\ce{3.00 L}$ container for $2$ mins, the total pressure in the container reached $\ce{6.50 atm}$. What is the partial pressure of the cyclobutane at this time?

(correct answer: 1.29 atm)

The first thing I noticed was that when the reaction is complete, there will be two moles of products for every one mole of products that existed prior to the reaction.

The first thing I tried doing was finding the concentration of cyclobutane at $t = 120\ce{\,s (2 min)}$. This was easy enough if we use the first-order integrated rate law: $$ \ce{[C4H8]_t = [C4H8]_0}\cdot e^{-kt} \\ \implies \ce{[C4H8]}_{t=120\text{ s}} = \left( \ce{\frac{2.26 mol}{3.00 L}} \right) \cdot e^{-(\ce{9.20\times 10^{-3} s^{-1}})(120\text{ s})} \\ = \ce{0.250 M} $$

But where do I go from here?

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2 Answers 2

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Found it.

So, after 2 minutes, the concentration of cyclobutane is around $0.250$ M, which means that around $0.750$ moles of cyclobutane is left in the reaction. If $0.750$ moles of cyclobutane are left, this means that $1.50$ moles of cyclobutane have reacted, so $2\times1.50 = 3$ moles of $\ce{C2H4}$ are left. So, at this time, a total of about $3.75$ moles of molecules, including both products and reactants, exist in the container.

Given this information, we can solve for the initial pressure of the gaseous mixture: $$\frac{P_{t = 0}}{2.26\text{ mol}} = \frac{6.50\text{ atm}}{3.75 \text{ mol}} \\ \implies P_{t = 0} = 3.92\text{ atm} $$

Finally, using the initial pressure, we can solve for cyclobutane's partial pressure at $t = 120$ seconds:

$$\frac{3.92\text{ atm}}{2.26\text{ mol}} = \frac{P_{t = 120 \text{ s}}}{0.75 \text{ mol}} \\ \implies P_{t = 120\text{ s}} = 1.30\text{ atm} $$

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The first thing OP has tried was the correct step, by finding the concentration of cyclobutane using the first-order rate law:

$$ \mathrm [A]_t = [A]_0 e^{-kt}$$

where $\mathrm [A]_t = [\ce{C4H8}]_t$ when $t = \pu{120 s}$, and $\mathrm [A]_0 = [\ce{C4H8}]_t = \pu{\frac{2.26}{3} mol L-1} = \pu{0.753 mol L-1}$ when $t = 0$.

Thus, at $t = \pu{120 s}$: $$ \ce{[C4H8]_t = [C4H8]_0}\cdot e^{-kt} \ \implies \ \ce{[C4H8]}_{t = \pu{120 s}} = \pu{0.753 mol L-1} \cdot e^{-(\pu{9.20 \times 10^{-3} s-1})(\pu {120 s})} \\ = \pu{0.250 mol L-1} $$

Therefore, the amount of $\ce{C4H8}$ remaining in the container is $\pu{0.250 mol L-1} \times \pu{3 L} = \pu{0.750 mol}$.

Meantime, above finding reveals that (according to the decomposition equation) $\pu{(2.26-0.750) mol}$ of $\ce{C4H8}$ has been decomposed to give $2 \times \pu{(2.26-0.750) mol = \pu{3.02 mol}}$ of $\ce{C2H4}$. Thus, the mole fraction of $\ce{C4H8}$ $(\chi_\ce{C4H8})$ in the container is: $$\chi_\ce{C4H8} = \frac{\pu{0.750 mol}}{\pu{(3.02 + 0.750) mol}} = 0.199$$ Thus, partial pressure of $\ce{C4H8}$ in the container $(p_\ce{C4H8})$ is: $$p_\ce{C4H8} = \chi_\ce{C4H8} \cdot P_\text{Tot} = 0.199 \cdot \pu{6.50 atm} = \pu{1.29 atm}$$

Note: I like the Mailbox's answer using the ratios. The reason to provide this answer is to show how to use the theory to solve the problem.

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    $\begingroup$ Thanks for posting this! I tried using mole fractions, but I ended up getting it wrong and I didn't know why. Seeing your answer helped me see why! $\endgroup$
    – Mailbox
    Mar 3, 2023 at 1:36

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