At 298 K, 1.35 mol BrCl(g) is introduced into a 12.0 L vessel, and equilibrium is established in the reaction.

Question

BrCl(g)-->0.5Br2(g) + 0.5Cl2(g)

Calculate the amount of BrCl(g) present when equilibrium is established.(Use delta G[Br2(g)]=3.11kJ/mol and delta G[BrCl(g)=-0.98Kj/mol)

I found Kp to be 0.359 using delta G = -RTlnK

Then I set up an ICE table where my equilibrium vaules were 1.35-x,0.5x and 0.5x and then I wrote my Kp expression and solved x = 0.849 and plugged this value into the x in my ice table. However I got the wrong answer. Does anyone know where I went wrong ?

Answer 1

I would venture the following:
1. I didn't get k = 0.359 but closer to 1
(1/2*3.11)-(-.98) =2.535
2.535 = -RTlnK. Did you use 8.314 for R?
2. Why do you call that Kp but work with moles? I suppose that doesn't matter since Kp = Kc. By amounts do they want moles?
3. I think what you did with the ICE chart is OK EXCEPT you did it in moles. USUALLY it works better in these cases to plug molar concns in instead of moles because we often forget that the answer comes out in moles and you must then convert to something else (unless of course by amounts they want mols).
4. What did you use for Kc? Was it
Kc = (Br2)^1/2(Cl2)^1/2/(BrCl). More than likely the K value and a faulty K expression (forgetting the sqrt rt is easy to do) are the problems. Let me know.

Answer 2

my K value was right. My mistake was when I plugged my values from my ICE table into the expression I used 0.25x^2/1.35-x instead of just x^2/1.35-x. Everything else was right.Thanks anyway

Answer 3

Thanks for letting me know. You're right on K. I redid mine and obtained 0.359. My mistake was not converting kJ to J before determining K.