Pure nitrosyl bromide, NOBr at 0.762 M is placed into a sealed vessel and heated up to 400 K. When the system reaches equilibrium, it is found that 65.1% of the nitrosyl bromide has decomposed to nitrogen monoxide and bromine. Determine Kc for this reaction at 400 K:

2 NOBr(g) -> 2 NO(g)+Br2(g)

If you start with 0.762M NOBr then NO will be 0.762 x 0.651 = about 0.496M and Br will be 1/2 that. The un-dissociated NOBr will be 0.762 x 0.349 = ?

Substitute those values into the Kc expression and solve.

To determine the equilibrium constant, Kc, for the given reaction, you need to first determine the concentrations of the reactants and products at equilibrium.

Let's assume that x is the extent of decomposition of NOBr (in moles) at equilibrium. Since the initial concentration of NOBr is 0.762 M, the concentration of NOBr at equilibrium will be (0.762 - x) M.

According to the balanced chemical equation, for every 2 moles of NOBr decomposed, 2 moles of NO and 1 mole of Br2 are formed. Therefore, the concentration of NO and Br2 at equilibrium will each be equal to x M.

The concentration of NOBr at equilibrium can be written as (0.762 - x) M, while the concentrations of NO and Br2 can be written as x M each.

Now, based on the given information, the extent of decomposition of NOBr is 65.1%. This means that 65.1% of NOBr has decomposed, which can be written as 0.651 times the initial concentration of NOBr: 0.651 * 0.762 M.

So, we can write the expression for the equilibrium constant, Kc, as follows:

Kc = [NO]^2 * [Br2] / [NOBr]^2

Substituting the equilibrium concentrations:

Kc = (x)^2 / [(0.762 - x)]^2

We know that x is equal to 0.651 * 0.762 M, so we can replace x in the equation:

Kc = (0.651 * 0.762)^2 / [(0.762 - 0.651 * 0.762)]^2

Now, you can plug in the values and calculate Kc.