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This is my reaction, $$\ce{Cd (g) + 1/2 Te2 (g) -> CdTe (s)}.$$ Considering the solid CdTe, $$K = P_{\ce{Cd}}^\mathrm{Eq}\cdot \left(P_{\ce{Te}}^\mathrm{Eq}\right)^{\frac{1}{2}}$$

I only have elementary knowledge on chemistry. I know that the equilibrium values of pressure are dependent on temperature. My question is - does increasing the partial pressure of one of the elements only changes the equilibrium pressure? If yes, for one or both?

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The equilibrium constant will stay the same at the same temperature. If the pressure of one component of the system in question is changed, equilibrium will be reestablished by either an change in the relative rates of the forward and reverse reactions.

Increasing the partial pressure of one reactant in this system will result in an increase in the rate of the forward reaction. Pressure of both both reactants will be reduced until equilibrium is reestablished.

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  • $\begingroup$ Thanks for your answer. I also sort of assumed the same, just wanted to verify with someone who knows these stuff. But the problem is, I am actually doing this experiment in the lab to deposit CdTe from elemental vapor. However, the observation doesn't match the theory. Increasing Te vapor increases the CdTe deposition rate, but increase of Cd vapor reduces the rate. Do you think, some kind of gas kinetics is involved here? $\endgroup$ – suhridkhan Feb 17 '15 at 4:34

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