Ap Chemistry

PCl3(g)+Cl2(g)⇄PCl5(g) Kc=0.11
A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl5(g) in the container is 0.040M.

On the graph above, carefully draw three curves, one for each of the three gases. The curves must show how the concentration of each of the three gases changes as equilibrium is established. Label each curve with the formula of the gas

(Can you just solve the problem so I can input it into the graph)

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  1. There isn't enough information to draw a graph. You have the concentrations at zero time and I can calculate the concentrations at equilibrium. There is no way to calculate the numbers between beginning and end because there is no data for that. To begin, the initial concentrations are 0.60 mol/L for PCl3 and 0.70 mol/L Cl2.
    ...............PCl3 + Cl2 ==> PCl5
    I.............0.60.......0.70..........0
    C................-x........-x............+x
    E...........0.60-x.....0.70-x........x
    Kc = 0.11 = (PCl5)/(PCl3)(Cl2)
    0.11 = (x)/(0.60-x)(0.70-x)
    Solve for x, then evaluate 0.60-x, and 0.70-x. That will give you the final (equilibrium) concentrations. You have the initial concentrations. I presume you just draw a straight line to connect the two point unless you want to try and fake a curved line but you have no point to support that fake line. Post your work on the solution above if you have trouble. Sorry I waited so long to answer. I didn't see it until just now. Good luck.

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    DrBob222

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