The reaction is Br2 (g) <---> 2Br (g).

It occurs at T = 1600 C.
When 1.05 mol Br2 are placed in a 2 Liter flask, 2.50% of Br2 undergoes dissociation. Calculate the Kp for the reaction.

Okay so i did the ICE table for this,
Br2 (g) <---> 2Br (g)
I : 1.05/2 ..... 0
C : -x ......... 2x
E : (1.05/2) - x .... 2x
But I don't know what to do next..
Please help and thank you in advance.

Where did the 1.05/2 come from? Does that reflect the 2.5%? If so shouldn't it be 1.05*0.025 = 0.02625?

......Br2(g) ==> 2Br(g)
I....1.05........0
C...-0.02625....2*(0.02625
E....1.024......0.0525

The E line uses mols; use PV = nRT to calculate p, the substitute into Kp expression and solve for Kp.

I've used too many significant figures so you need to go through the estimates I've made with that in mind.

Next, you can use the given information that 2.50% of Br2 dissociates to calculate the equilibrium concentrations.

Since 2.50% of Br2 dissociates, it means that 2.50/100 * 1.05/2 moles of Br2 dissociates, which is equal to 0.02625 moles of Br2.

Using the equation from the ICE table, you can set up the expression for equilibrium constant Kp:

Kp = [Br]^2 / [Br2]

Where [Br] represents the concentration of Br and [Br2] represents the concentration of Br2.

Substituting the equilibrium concentrations into the equation, we have:

Kp = (2x)^2 / ((1.05/2) - x)

Now you can substitute the value of x, which represents the equilibrium concentration of Br2, using the value calculated earlier.

Kp = (2 * 0.02625)^2 / ((1.05/2) - 0.02625)

Simplifying the equation will give you the value of Kp.

To find the value of Kp for the given reaction, we can use the expression:

Kp = (P_Br)^2 / P_Br2

Where P_Br and P_Br2 represent the partial pressures of Br and Br2, respectively.

From the ICE table, we can determine that the equilibrium concentration of Br2 is (1.05/2) - x and the equilibrium concentration of Br is 2x. However, since the question gives the percentage dissociation of Br2, we can directly determine the equilibrium concentration of Br2 as 0.0250(1.05) mol/liter (as 2.50% of Br2 dissociates).

So, [Br2] (equilibrium) = 0.0250(1.05)/2 M
= 0.013125 M

Now, as the reaction occurs in a 2-liter flask, the partial pressure of Br2 (P_Br2) is given by:

P_Br2 = (0.013125 M) (0.08206 L.atm/mol.K) (1600+273.15 K)
= 4.237 atm

The partial pressure of Br (P_Br) is given by:

P_Br = (2)(0.013125 M) (0.08206 L.atm/mol.K) (1600+273.15 K)
= 8.474 atm

Now, we can calculate the value of Kp by substituting these values into the expression:

Kp = (P_Br)^2 / P_Br2
= (8.474 atm)^2 / 4.237 atm
= 16.95 atm

Therefore, the value of Kp for the given reaction at T = 1600°C is 16.95 atm.