0
$\begingroup$

Question

A palladium or platinum catalyst was used in an automobile to convert carbon monoxide gas to carbon dioxide according to the following reaction:

$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$

A chemist researching the effectiveness of a new catalyst combines a 2.0 : 1.0 mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45 L flask at a total pressure of 745 torr and a temperature of 552 K. When the reaction is complete, the pressure in the flask has dropped to 552 torr.

What percentage of the carbon monoxide was converted to carbon dioxide?

(a) 67.7 %

(b) 85.7 %

(c) 77.5 %

(d) 57.8 %

(e) 46.5 %


My attempt

I calculated the moles of gas in the original system

$$n = \frac{PV}{RT} = \frac{745 \times 2.45}{8.31 \times 552} = 0.398 \, \mathrm{mol}$$

Then I worked out the moles of $\ce{CO}$ and $\ce{O2}$ in the original system using the mole fractions.

$$\ce{2CO + O2 -> 2CO2}$$

moles of $\ce{CO}$ = $(2/3) \times 0.398 = 0.265\, \mathrm{mol}$

moles of $\ce{O2} = (1/3) \times 0.398 = 0.1326\,\mathrm{mol}$

moles of $\ce{CO2} = 0\, \mathrm{mol}$

Then I calculated the number of moles in the new system.

$$n-\frac{PV}{RT} = \frac{552 \times 2.45}{8.31 \times 552} = 0.295\, \mathrm{mol}$$

Then I worked out the moles of $\ce{CO, O2}$ and $\ce{CO2}$ in the new system using the mole fractions.

$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$

moles of $\ce{CO} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$

moles of $\ce{O2} = (1/5) \times 0.295 = 0.059\, \mathrm{mol}$

moles of $\ce{CO2} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$

Therefore percentage of CO reacted

$$\%_{\ce{CO}} = \frac{0.265 - 0.118}{0.265} \times 100 = 55.5\, \%$$

The correct answer according to the mark scheme is C) 77.5%. Is there any other method to get this answer?

$\endgroup$
1
  • 3
    $\begingroup$ You need not to compute molar amounts at all. You can compute the final/initial pressure ratio that is equal to final/initial molar amount ratio. From that, you can directly calculate the percentage of conversion from inverted function pressure ratio = f(conversion percentage). $\endgroup$
    – Poutnik
    Jan 16, 2023 at 9:10

2 Answers 2

3
$\begingroup$

Using the ideal gas state equation as the starting point:

$pV=nRT \implies \frac{p_1V}{p_2V}=\frac{n_1RT}{n_2RT} \\ \implies \frac{p_1}{p_2}=\frac{n_1}{n_2}$

leading to the consequence that the gas pressure is proportional to the total molar amount of present gases.

As there is molar conversion in ratio 2/3, the final/initial pressure ratio 2/3 would mean 100 % conversion.

The rest below is just interpolation between the ratio 1 (0 %) and 2/3 (100 %)

If

  • $\alpha$ is the percentual conversion
  • $p_\mathrm{tot,ini}$ is the initial pressure of gases
  • $p_\mathrm{tot,fin}$ is the final pressure of gases

then

$$\frac{p_\mathrm{tot,fin} }{ p_\mathrm{tot,ini}} = 1 - \frac 13 \cdot\frac{\alpha}{100}$$

$$\alpha = (1 - \frac{p_\mathrm{tot,fin} }{ p_\mathrm{tot,ini}})\cdot 300 \ \%= (1 - \frac{\pu{552 torr} }{ \pu{745 torr}})\cdot 300 \ \% \approx 77.7\ \% $$

$\endgroup$
2
$\begingroup$

Let:

A represent $\ce{CO}$

B represent $\ce{O2}$

C represent $\ce{CO2}$

So the reaction becomes:

$$\ce{2A + B->2C}$$

We begin by calculating the initial molar fraction of A.

Since only A and B are initially present:

$$y_{Ao}+y_{Bo}=1$$

Since the initial molar ratio between both species is 2:1, then:

$$y_{Ao}=2\;y_{Bo}$$

Using both equations to solve for $y_{Ao}$, we get:

$$y_{Ao}=\frac{2}{3}$$

Next, we observe that between the initial and final state, temperature and volume are constant, but there is a change in moles between reactants and products, so:

$$T=T_o$$

$$V=V_o$$

$$\Delta n=-1$$

If we divide the ideal gas law in the final state by the initial state, we get:

$$\frac{PV}{P_oV_o}=\frac{nRT}{n_oRT_o}$$

Cancelling all equal terms:

$$\frac{P}{P_o}=\frac{n}{n_o}$$

The molar ratio on the right hand can be expressed in terms of conversion $X$ and expansion coefficient $\epsilon$ after performing a molar balance of A, B, and C:

$$ \require{cancel} \begin{align} n_A=n_{Ao}-ax \\ n_B=n_{Bo}-bx \\ n_C=n_{Co}+cx \\ \hline n=n_o+\Delta nx \end{align} $$

In this case, $\Delta n=c-a-b$, and small $x$ represents the non-normalized conversion with respect to the limiting reagent. In order to normalize it, we arbitrarily define A as the limiting reagent, and its conversion as:

$$X=\frac{ax}{n_{Ao}}$$

Solving for $x$:

$$x=\frac{n_{Ao}X}{a}$$

Substituting $x$ into the overall molar balance:

$$n=n_o+n_{Ao}\frac{\Delta n}{a}X$$

Dividing both sides by $n_o$:

$$\frac{n}{n_o}=1+\frac{n_{Ao}}{n_o}\frac{\Delta n}{a}X$$

Or equivalently:

$$\frac{n}{n_o}=1+y_{Ao}\frac{\Delta n}{a}X$$

Then, we define:

$$\epsilon=y_{Ao}\frac{\Delta n}{a}$$

And substitute it above to get:

$$\frac{n}{n_o}=1+\epsilon X$$

Which can also be substituted above:

$$\frac{P}{P_o}=1+\epsilon X$$

Solving for $X$:

$$X=\frac{\frac{P}{P_o}-1}{\epsilon}$$

Calculating $\epsilon$:

$$\epsilon=y_{Ao}\frac{\Delta n}{a}=\frac{2}{3}\;\frac{-1}{2}=-\frac{1}{3}$$

Plugging in all numerical values:

$$X=\frac{\frac{552}{745}-1}{-\frac{1}{3}}$$

So we get:

$$X=0.777$$

Or equivalently 77.7%

$\endgroup$
12
  • $\begingroup$ Hmm, 1-2 equations should be enough. $\endgroup$
    – Poutnik
    Jan 16, 2023 at 19:44
  • $\begingroup$ @Poutnik Elaborate answers is my style :) or at least in cases where it's likely the level implicated is high school/college. $\endgroup$
    – Sam202
    Jan 16, 2023 at 19:50
  • 2
    $\begingroup$ After some point it may become intimidating, like more complex than it is. But yes, I know your style. :-) $\endgroup$
    – Poutnik
    Jan 16, 2023 at 19:52
  • $\begingroup$ @Sam202 This is great. Thank you so much! $\endgroup$
    – Jane902
    Jan 17, 2023 at 11:42
  • $\begingroup$ @Sam202 However this step "The molar ratio on the right hand can be expressed in terms of conversion X and expansion coefficient e" is not something taught in high school chemistry or biology. Can you please expand on that or direct me to a resource that explains where the expansion coefficient comes from for a bio student? $\endgroup$
    – Jane902
    Jan 17, 2023 at 12:03

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.