At 298 K, 1.35 mol BrCl(g) is introduced into a 12.0 L vessel, and equilibrium is established in the reaction.
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 ?
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.
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
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.
It seems like you are on the right track with setting up an ICE table and using the Kp expression to solve for the value of x. However, there might be a mistake in your calculation or use of units. Let me explain the correct approach in more detail.
First, let's write the balanced equation for the reaction:
BrCl(g) --> 0.5Br2(g) + 0.5Cl2(g)
Based on this equation, the stoichiometric coefficients tell us that for every 2 moles of BrCl(g) that react, we will form 1 mole each of Br2(g) and Cl2(g).
Next, let's define the initial and equilibrium concentrations of BrCl(g) as [BrCl]_0 and [BrCl], respectively. We know that initially, 1.35 mol of BrCl(g) is introduced into the 12.0 L vessel, so the initial concentration ([BrCl]_0) can be calculated as:
[BrCl]_0 = moles/volume = 1.35 mol / 12.0 L = 0.1125 M
Now, we set up the ICE table:
BrCl(g) --> 0.5Br2(g) + 0.5Cl2(g)
Initial: 0.1125 M 0 M 0 M
Change: -x +0.5x +0.5x
Equilibrium: 0.1125 - x 0.5x 0.5x
According to the balanced equation, the equilibrium concentration of BrCl is 0.1125 M - x. The equilibrium concentrations of Br2 and Cl2 are both 0.5x.
Now, let's substitute these values into the Kp expression:
Kp = (P[Br2]^(0.5) * P[Cl2]^(0.5)) / P[BrCl]
Since the gases are in the same container, we can assume that their partial pressures are proportional to their concentrations. Therefore, we have:
Kp = ([Br2]^(0.5) * [Cl2]^(0.5)) / [BrCl]
Since [Br2] and [Cl2] are both 0.5x and [BrCl] is 0.1125 - x, we can express the Kp value as:
Kp = ((0.5x)^(0.5) * (0.5x)^(0.5)) / (0.1125 - x)
Now, substitute the given value of Kp (0.359) into the equation:
0.359 = (0.5x)^(0.5) * (0.5x)^(0.5) / (0.1125 - x)
To solve this equation for x, you can square both sides and rearrange to isolate x. However, keep in mind that this is a quadratic equation, so there might be two possible solutions (one positive and one negative). Make sure to choose the appropriate solution that falls within the range of physically meaningful values (0 to 1.35 mol).
Once you find the value of x, you can calculate the equilibrium concentration of BrCl as (0.1125 - x).