At 500K,1.00mol of ONCl is introduced into 2.50dm3 closed vessel.At equilibrium the ONCl is 6.5%dissoiated.What is the value of the Kc for equilibrium at 500K

To determine the value of Kc for the equilibrium at 500K, we need to use the given information about the initial amount of ONCl and the percentage of dissociation at equilibrium.

First, let's calculate the number of moles of ONCl that dissociated at equilibrium.

The initial number of moles of ONCl is given as 1.00 mol. Since the percentage of dissociation is given as 6.5% (or 0.065), we can calculate the number of moles dissociated as follows:

moles dissociated = 1.00 mol * 0.065 = 0.065 mol

Next, let's calculate the number of moles remaining at equilibrium. Since the ONCl is 6.5% dissociated, the remaining amount is:

moles remaining = 1.00 mol - 0.065 mol = 0.935 mol

Now, we can use the equation for Kc to find its value. The equation for the dissociation of ONCl can be represented as follows:

ONCl ⇌ O + NCl

The balanced equation shows that for every 1 mol of ONCl, 1 mol of O and 1 mol of NCl are formed. Therefore, at equilibrium, the concentration of O and NCl is equal to the moles remaining (0.935 mol).

Using the given volume of the closed vessel (2.50 dm3 or 2.50 L), we can convert the moles to molarity (M):

Molarity = moles remaining / volume

Molarity = 0.935 mol / 2.50 L = 0.374 M

Since the concentrations of O and NCl are equal, we can use this value (0.374 M) for both in the Kc expression:

Kc = [O][NCl] / [ONCl]

Kc = (0.374)(0.374) / 0.065

Kc = 2.148

Therefore, the value of Kc for the equilibrium at 500K is approximately 2.148.