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Suppose $K=4.5 \times 10^{-3}$ at a certain temperature for the reaction$$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$If it is found that the concentration of $\mathrm{PCl}_{5}$ is twice the concentration of $\mathrm{PCl}_{3},$ what must be the concentration of Cl, under these conditions?

$\left[\mathrm{Cl}_{2}\right]=9.0 \times 10^{-3} \mathrm{M}$

Chemistry 102

Chapter 17

Equilibrium

Section 9

Solubility Equilibria

Chemical Equilibrium

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this question is dealing with this equilibrium. We have PCL five as a gas being an equilibrium with PCL three as a gas and seal, too. This actually really common example that textbooks use for an equilibrium of gas. And I'm not sure why, but usedto. Either way, we're given the K value for this, which, as always, is an equilibrium. Positive describes any quick him. And if you go to 4.5 times 10 to the negative 30. And so we're told that an equilibrium we have two times as much PCL five as you have PCL three. So let's just generically pick some concentrations. I'll give that. Let's say we have to Moller for PCL five and one Moeller her PCL three and then we're still trying to find what seal to is that's were asked to do is and so just say, acts there. And so we actually can easily solve this. We just had to write out what kay is in terms of the mass action expression. So remember, a mass act expression means the concentration of all the products of PCL three times a concentration of seal, too, over the concentration of the reactions PCL five. It's Now we plug in these values that we've arbitrarily given. We have PCL three, we have one times X over two coming from here. And so when you solved this so month by two times times two, that's gonna equal nine times. Tencent, your third. And that's our concentration. Four C l two. Um, remember, even though K is a universe value the concentration of steel to has a concentration units of polarity and so put that in right there.

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