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The below equilibrium reaction takes place among the burnt exhaust gases still left inside the engine cylinder and being reused again for combustion. The author has specified the equilibrium between only three of the exhaust gases. I’ve been asked to find the final pressure and degree of dissociation.

QUESTION: Say I have the following equilibrium reaction $$\ce{CO + 1/2 O2 <=> CO2 }$$ The stoichiometric mixture of $\ce{CO}$ and $\ce{O2}$ in a closed vessel, initially at 1 atm and 300 K, is exploded. Calculate the composition of the products of combustion at 2500 K and the gas pressure. Take $K_p=27.5$. Take $\alpha$ as the degree of dissociation.

MY ATTEMPT AT AN ANSWER: I use the ICE method to find out the reactant and product composition $$\ce{CO + 1/2 O2 <=> CO2}$$ The reaction quotient can be given as $Q=0/0.5=0$. Therefore, the products must have plus sign.

Then I get the final partial pressures (NOT composition(/concentration)) as $$\ce{CO}=[1-\alpha]$$ $$\ce{O2}=[1/2-\alpha/2]$$ $$\ce{CO2}=[\alpha]$$

then, $$ K_p= \frac{\alpha}{[1-\alpha][1/2-\alpha/2]^{0.5} } $$

i.e $$K_p= \frac{\alpha \sqrt2}{[1-\alpha]^{3/2} } $$

Solving for $\alpha$, I get $\alpha=0.87$

I would then use the IDEAL GAS LAW to find the product mixture, i.e $$p_\text{r}V=n_\text{r}RT_\text{r}$$ $$p_\text{p}V=n_\text{p}RT_\text{p}$$

I would then substitute these values into the expression for $K_p$.

MY QUERY: 1.Is my calculation for $K_p$ correct? I take $\ce{CO2}$ as the product and $\ce{CO}$ and $\ce{O2}$ as the reactants.

2.I am not able to figure out what is the amount of substance of the products $n_\text{p}$ and the amount of substance of the reactants $n_\text{r}$. The book's solution gives $n_\text{r}=3/2$ and $n_\text{p}=1+\alpha/2$.

The author solves with the initial species as $\ce{CO2}$ and obtains the compositions

$\ce{CO2}=(1-\alpha);\quad\ce{CO}=\alpha;\quad\ce{O2}=\alpha/2$, which is confusing given that the question states the reaction starts with $\ce{CO}$ and $\ce{O2}$ in a closed vessel.

Thank you. I solve for $\alpha$ and obtain a value of 0.87, it was a cubic polynomial.

I am only able to obtain the amount of substance per unit volume, but not the amount of substance as such (WHY?).

My calculation is as follows
I know that the total partial pressure can be given as $p=\frac{3}{2}-\alpha/2$, i.e taking the initial reactant partial pressure as 1 atm.

Then, $p_\text{r}+p_\text{p}=3/2-\alpha/2$ where, $p_\text{r}=3/2-\frac{3}{2}\alpha/2$ and $p_\text{p}=\alpha$

therefore, $\frac{n_\text{r}RT_\text{r}}{V}+ \frac{n_\text{p}RT_\text{p}}{V}=3/2+\alpha/2$

$\frac{R}{V}(n_\text{r}T_\text{r}+n_\text{p}T_\text{p})=3/2+\alpha/2\tag{eq1}$

$\frac{p_\text{r}}{p_\text{p}}=\frac{n_\text{r}T_\text{r}}{n_\text{p}T_\text{p}}\tag{eq2}$

solving for $n_\text{p}$ and $n_\text{r}$, I get $n_\text{p}/V=3.24\times 10^{-3}$ and $n_\text{r}/V=0.025$

Example 3.4 is the problem I'm having trouble with:

enter image description here

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  • $\begingroup$ Your equations for the final partial pressures are not correct because the final temperature is much larger than the initial one. The equations are only correct as long as both volume and temperature remain constant (the same at initial and final time). $\endgroup$ – perplexity Oct 7 '13 at 11:07
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I would say the biggest issue in your approach is in confusing $K_c$ and $K_p$. You listed your (indeed, correct) equations for the products and reactants in terms of alpha, but you cited them as concentrations instead of partial pressures.

You are most of the way there, after plugging in your $K_p$ value to solve for alpha (which is the composition), use the expressions you wrote above for the partial pressures of each gas and use the result in the ideal gas law to find moles.

EDIT: I apologize for the mistake, I didn't read the question carefully enough.

You actually do have the volumes...sort of. It's a closed container, so you know that the volume doesn't change. Since you are only looking for pressures you can use the form:

$\left(\dfrac{p}{nT}\right)_\text{initial} = \left(\dfrac{p}{nT}\right)_\text{final}$

You know the initial pressures of each reactant (fractions of the total pressure by molar ratio), the temperatures (given), and the moles (found from $K_p$). Hope that helps!

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  • $\begingroup$ I have edited my answer for the number of moles, but I am still not able to eliminate the volume. What other equation am I missing? $\endgroup$ – Sunny Marella Jun 12 '13 at 5:16
  • $\begingroup$ Sorry for not reading the question, I'll add an edit $\endgroup$ – Gavin Kramar Jun 12 '13 at 14:33
  • $\begingroup$ Did you mean instead:(p/nT)Reactant=(p/nT)product? $\endgroup$ – Sunny Marella Jul 3 '13 at 12:15
  • $\begingroup$ To clarify, that sloppy notation was supposed to indicate that the pressure divided by the moles and temperature of the reactants is equal to the corresponding values of the products. $\endgroup$ – Gavin Kramar Jul 5 '13 at 16:36
  • $\begingroup$ Please see attached the exercise problem I'm struggling with. The author has arrived at the values for K_p from the temperatures emperically. But,what I do not understand is what he equates it with in page 89. Help!! $\endgroup$ – Sunny Marella Jul 6 '13 at 15:45

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