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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 ...

2

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 ...

3

The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$ The additional pressure compresses the liquid and thereby increases its chemical ...

1

You can relate an equilibrium constant $K_p$ written in terms of partial pressures to one written in terms of mole fractions $K_\chi$, as follows: $$K_\chi=K_p (p/p_0)^{-\Delta \nu}\tag{1}$$ where $\Delta \nu$ is the difference in the stoichiometric coefficients for a molar amount of reaction. $K_p$ is independent of the pressure $p$ of the system but ...

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For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant. When pressure is decreased($P_{total}$), \$ ...

2

There are a number of ways to look at this. One is to look at the equations. Another is to plot the pressure p as a function of volume V. The area under the pressure curve on the plot is the total work done on the system. It is very easy to answer your second question by inspection of this figure, but perhaps not the first question without actually ...

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