$\ce{ZnO}$ is exposed to pure $\ce{CO}$ at $\pu{1300 K}$ and the equilibrium $$\ce{ZnO(s) + CO(g) <=> Zn(g) + CO2(g)}$$

is then established at $\pu{1 atm}$ pressure. The density of the gas mixture is $\pu{0.344 g L-1}$ at the same temperature. Calculate the partial pressure of $\ce{CO(g)}$ at equilibrium (atomic weight of $\ce{Zn}$ is $65.4).$

My attempt

Since at equilibrium the amounts of $\ce{CO}$, $\ce{Zn}$ and $\ce{CO2}$ are equal (as no information about initial amounts are given), mole fraction of $\ce{CO}$ is 1/3.

Since pressure is given as $\pu{1 atm},$ the partial pressure of $\ce{CO}$ is 1/3 atm. Where I am wrong?

  • $\begingroup$ You might want to revise your idea of equilibrium. $\endgroup$ – Ivan Neretin Oct 17 '19 at 15:27
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    $\begingroup$ @GRSousaJr: What about $\ce{Zn}$ vapors? At $\pu{1600 K}$, $\ce{Zn}$ stays as vapors (or gas). $\endgroup$ – Mathew Mahindaratne Oct 17 '19 at 16:37
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    $\begingroup$ @MathewMahindaratne Sorry! My mistake! $\endgroup$ – GRSousaJr Oct 17 '19 at 16:46

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:

$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$

Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:

$$p_1 = x_1p\tag{1}$$

To find unknown mole fraction, two facts can be taken into consideration:

  1. As Karsten also pointed out, the molar ratio between the products is $1:1$ so that $x_2 = x_3$. Which gives

    $$x_1 + x_2 + x_3 = 1 \quad\implies\quad x_2 = x_3 = \frac{1 - x_1}{2}\label{eqn:2}\tag{2}$$

  2. All three mole fractions are linked together in the expression for the average molar mass:

    $$\bar{M} = x_1M_1 + x_2M_2 + x_3M_3\label{eqn:3}\tag{3}$$

The average molar mass can be found via average density of the gaseous mix $\bar{ρ}$ by applying ideal gas law:

$$pV = nRT$$

$$p = \frac{nRT}{V} = \frac{mRT}{\bar{M}V} = \frac{\bar{ρ}RT}{\bar{M}}$$

$$ \begin{align} \bar{M} &= \frac{\bar{ρ}RT}{p} \\ &= \frac{\pu{0.344 g L-1} × \pu{0.082 L atm K-1 mol-1} × \pu{1300 K}}{\pu{1 atm}} \\ &\approx \pu{36.67 g mol-1}\tag{4} \end{align} $$

Now, knowing numerical value for $\bar{M},$ we can combine \eqref{eqn:2} and \eqref{eqn:3}:

$$\bar{M} = x_1M_1 + \frac{1 - x_1}{2}M_2 + \frac{1 - x_1}{2}M_3$$

Grouping the like terms and expressing $x_1$:

$$ \begin{align} x_1 &= \frac{M_2 + M_3 - 2\bar{M}}{M_2 + M_3 - 2M_1} \\ &= \frac{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{36.67 g mol-1}}{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{28.01 g mol-1}} \\ &\approx 0.67\tag{5} \end{align} $$

Finally, the partial pressure of carbon monoxide:

$$p_1 = 0.67×\pu{1 atm} = \pu{0.67 atm}$$


Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given)

If no information is given, you should just give the three amounts names and treat them as unknowns.

ZnO is exposed to pure CO at 1300 K

This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental zinc will be present in a 1:1 ratio because that is the ratio they are produced at.

The density of the gas mixture is 0.344 g/L at the same temperature.

Let's find the density of pure CO at 1300 K and 1 atm pressure, assuming it behaves as ideal gas.

$$\rho_1 = m/V = M n / V = M P / (R T) = $$ $$\frac{\pu{28.01 g/mol} \cdot \pu{1 atm}}{\pu{0.08206 L atm K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.2626 g/L}$$

And now the density of a 1:1 mixture of carbon dioxide and zinc under the same conditions. The average molar mass of the two species is 0.5 (65.4 + 44.01) g/mol.

$$\rho_2 = m/V = M_\mathrm{ave} n / V = M P / (R T) = $$ $$\frac{\pu{54.71 g/mol} \cdot \pu{1 atm}}{\pu{8.314 J K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.5129 gm3 g/L}$$

So the density of the gas for pure reactants is lower than the observed density, and the density for pure products is higher. The equilibrium mixture contains some reactant and some product, and the density is a weighted average of the two densities just calculated (with $X_\ce{CO}$ the mole fraction of carbon monoxide in the gas phase of the reaction mixture):

$$\rho = X_\ce{CO} \rho_1 + (1-X_\ce{CO}) \rho_2$$

We can solve for $X_\ce{CO}$ and calculate its value:

$$\rho - \rho_2 = X_\ce{CO} (\rho_1 - \rho_2)$$

$$ X_\ce{CO} = \frac{\rho - \rho_2}{\rho_1 - \rho_2} = 0.675 $$

So the partial pressure of carbon monoxide is 0.675 atm.

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    $\begingroup$ How to do this without percent completion of reaction as we aren't taught that. Moreover, I think R should be taken as 0.082 L atm K-1 to match the units. And the answer key given to this question was 2/3. $\endgroup$ – Equation_Charmer Oct 18 '19 at 0:50
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    $\begingroup$ I'm not sure it's correct to write "$\rho = \xi \rho_2 + (1-\xi) \rho_1$" for density. $\endgroup$ – andselisk Oct 18 '19 at 10:38
  • $\begingroup$ @andselisk - There were two problems in my answer. I mislabeled what I called $\xi$ as extent of reaction (it is just a mole fraction), and then I gave the partial pressure of the (combined) products rather than the reactants. I obviously confused myself with my own choice of variables, so in the corrected answer I simply used the mole fraction of carbon monoxide to express the density of the gas mixture. Thanks for pointing it out. $\endgroup$ – Karsten Theis Oct 18 '19 at 11:31
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    $\begingroup$ @dhaneshvijay I fixed my answer after looking at andselisk's, who got the correct result that matches what you saw in the answer key. I also swapped the units for R so it is easier to follow the arithmetic. $\endgroup$ – Karsten Theis Oct 18 '19 at 11:33

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