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?


2 Answers 2


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:

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


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.