# Equilibrium reaction ICE method to ideal gas law

1. Jun 5, 2013

### marellasunny

QUESTION: Say I have the following equilibrium reaction
$$CO+\frac{1}{2}O_2\leftrightharpoons CO_2$$
The stoichiometric mixture of CO and $O_2$ in a closed vessel, initially at 1 atm and 300K, is exploded. Calculate the composition of the products of combustion at 2500K and the gas pressure.
Take $K_p$=27.5. Take $\alpha$ as the degree of dissociation.

I use the ICE method to find out the reactant and product composition
$$CO+\frac{1}{2}O_2\leftrightharpoons CO_2$$
The reaction quotient can be given as Q=0/0.5=0. Therefore, the products must have plus sign.

Then I get the final composition(/concentration) as
$$CO=[1-\alpha]$$
$$O_2=[1-\alpha/2]$$
$$CO_2=[\alpha]$$

then, $$K_p= \frac{\alpha}{[1-\alpha][1-\alpha/2]^{0.5} }$$
I would then use the IDEAL GAS LAW to find the product mixture

i.e $$p_RV=n_RRT_R$$ $$p_pV=n_PRT_p$$

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

MY QUERY:
1.Is my calculation for $K_P$ correct? I take $CO_2$ as the product and CO and $O_2$ as the reactants.

2.I am not able to figure out what is the number of moles of the products $n_P$ and the number of moles of the reactants $n_R$.The book's solution gives $n_R$=3/2 and $n_P=1+\alpha/2$.

The author solves with the initial species as $CO_2$ and obtains the compositions $CO_2=(1-\alpha)$;$CO=\alpha$;$O_2=\alpha/2$, which is confusing given that the question states the reaction starts with CO and O2 in a closed vessel.

2. Jun 5, 2013

### Staff: Mentor

If the mixture was stoichiometric, initial amount of oxygen was not 1. Unless I am missing something.

See above. Other than that looks like you are on the right track.

Actually initial composition doesn't matter (as long as it is stoichiometric), as if the mass balance doesn't change, final equilibrium will be identical. Calling α degree of dissociation suggests CO2 and its decomposition as a starting point. Doesn't mean your approach is wrong, it should yield the same result in terms of final pressures. But as α's are different, final result expressed using your α and using book's α won't look identical.

3. Jul 6, 2013

### marellasunny

Borek,please find attached the problem I'm struggling with. I've highlighted the step which I do not understand also.

The K_p values were arrived from the temperatures empirically. But,what is the expression to the left in page 89?

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4. Jul 7, 2013

### Staff: Mentor

Hard to say something specific without seeing eq 3.40. But in general it looks like equilibrium expressed using reaction degree (fraction) - as defined at the very top of the page. In a way similar to http://en.wikipedia.org/wiki/Ostwald_dilution_law

5. Jul 7, 2013

### marellasunny

Oswald Dilution Law, very helpful insight.Since, concentration=n/V , this is not how it is expressed in the red highlighted area(attachment previous post). I first thought this was a relation between K_p and K_c,but the units didn't agree. It doesn't agree even in the Oswald dilution law.
$$\frac{1-\alpha }{\alpha (\alpha /2)^{0.5}}{\frac{n_p}{p_p}^{0.5}}=K_p$$
I also Attached 3.40 eq.

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