The compound pcl5 decomposes into cl2 and pcl3. the equilibrium of pcl5(g) cl2(g) pcl3(g) has a keq of 2.24 x 10-2 at 327°c. what is the equilibrium concentration of cl2 in a 1.00 liter vessel containing 0.235 mole of pcl5 and 0.174 mole of pcl3? remember to use the correct number of significant digits

Answers:


    your intermediate answer, without proper significant digits, is 0.030253 mol.

    rounding with the 5 even rule, the answer is  0.0302 mol.
    ignoring the 5 even rule, the answer is  0.0303 mol.



    pcl₅(g) ⇄ cl₂(g) + pcl₃

so for this equilibrium:

   

the vessel being 1.00 l means that the moles of each compound is the same as the molarity.

thus our current information is as follows

    [pcl₅] = 0.235 m
    [pcl₃] = 0.174
    keq = 2.24  × 10⁻²

we need to find [cl₂].
rearranging the equilibrium expression to isolate [cl₂]:

   

substituting in our information, we get

   

your intermediate answer is 0.030253 mol.
rounding with the 5 even rule, the answer is  0.0302 mol.
ignoring the 5 even rule, the answer is  0.0303 mol.