Molecular bromine is 24 percent dissociated at 1600 K and 1.00 bar in the equilibrium Br2 (g) = 2Br (g).

Calculate Kp at 1600 K.

Kp = (2Br)^2/(Br2) = 4

To calculate Kp at 1600 K, we need to use the expression Kp = (PBr)^2 / PBr2, where PBr is the partial pressure of Br and PBr2 is the partial pressure of Br2.

Given that molecular bromine (Br2) is 24 percent dissociated, it means that the concentration of Br2 decreases by 24 percent, and the concentration of Br increases by 24 percent. This implies that the concentration of Br2 is only 76 percent of its original value, and the concentration of Br is 124 percent of its original value.

Assuming that the initial pressure of Br2 is 1.00 bar, after dissociation, the partial pressure of Br2 is 0.76 bar (76 percent of 1.00), and the partial pressure of Br is 1.24 bar (124 percent of 1.00).

Now we can substitute these values into the Kp expression: Kp = (PBr)^2 / PBr2 = (1.24)^2 / 0.76.

Calculating this, we get Kp = 1.6026.

Therefore, the value of Kp at 1600 K is approximately 1.6026.

To calculate the equilibrium constant Kp at a given temperature, we need to know the degree of dissociation (α) of the molecule.

Given that molecular bromine (Br2) is 24 percent dissociated, we can assume that α = 0.24.

The equilibrium equation is: Br2 (g) = 2Br (g)

Using the equilibrium expression in terms of partial pressures: Kp = (P_Br)^2 / P_Br2

To calculate Kp, we need to determine the partial pressures of Br and Br2.

Since Br2 is 24 percent dissociated, the equilibrium concentration of Br2 would be 1 - α = 1 - 0.24 = 0.76.

Therefore, the partial pressure of Br2 (P_Br2) would be 0.76 * 1.00 bar = 0.76 bar.

Since Br2 dissociates to form 2 moles of Br, the concentration of Br would be twice the concentration of Br2: 2 * 0.24 = 0.48.

Therefore, the partial pressure of Br (P_Br) would be 0.48 * 1.00 bar = 0.48 bar.

Now we can substitute these values into the equilibrium expression: Kp = (0.48)^2 / 0.76 = 0.307

Therefore, the equilibrium constant Kp at 1600 K is 0.307.