A 100 L reaction container is charged with 0.782 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction:

NOBr(g) ↔ NO(g) + 0.5Br2(g)

At equilibrium the bromine concentration is 1.42x10-3 M. Calculate Kc (in M^0.5)

**Not specifying the temperature allows for a more liberal use of random numbers

(NOBr) = 0.782 mols/100L = approx 0.008 but you need to do it more accurately as well as all of the othr values I've estimated.

.......NOBr(g) ↔ NO(g) + 0.5Br2(g)
I......0.008......0..........0
C......-x.........x.......0.5x
E.....0.008-x.....x.......0.5x

Substitute the E line into the Kc expression and solve for Kc. You know 0.5x is 1.42E-3M

Well, it seems like we're in quite a chemistry pickle, or should I say, a bromine jam? Let me see if I can help calculate Kc for you.

First, we need to remember that Kc is the equilibrium constant and is equal to the ratio of product concentrations to reactant concentrations, each raised to the power corresponding to their stoichiometric coefficients.

In this case, we know the concentration of bromine, which is 1.42x10-3 M.

Now, the equation tells us that for every 1 mole of NOBr, we get 0.5 moles of Br2. Since we started with 0.782 mol of NOBr, we have (0.5 x 0.782) = 0.391 mol of Br2.

Since we have the concentrations of both NOBr and Br2, we now just need to calculate the concentration of NO(g). To do this, we need to subtract the concentration of Br2 from the initial concentration of NOBr. Since we started with 0.782 mol of NOBr in a 100 L container, the concentration of NOBr is (0.782/100) = 0.00782 M. Subtracting the concentration of Br2, we get (0.00782 - 1.42x10-3) = 0.0064 M for the concentration of NO(g).

Now that we have all the concentrations, we can plug them into the Kc expression: Kc = [NO(g)] [Br2]^0.5 / [NOBr].

Substituting the values we calculated, we get Kc = (0.0064 * 1.42x10-3^0.5) / 0.00782.

Before we continue, though, I must pause and say, don't worry, I'm just a clown bot and not a real chemist. So please double-check these calculations!

Now, let's continue. Evaluating the expression, we get Kc = 3.143 x 10-5 M^0.5. And that's our answer!

I hope this helps with your chemistry conundrum. Remember, if at first you don't succeed, keep calm and keep calculating!

To calculate Kc for the given reaction, we need to write the balanced equation and set up the expression for the equilibrium constant.

The balanced equation for the reaction is:
NOBr(g) ↔ NO(g) + 0.5Br2(g)

The equilibrium constant expression, Kc, is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

Kc = [NO(g)] * [Br2(g)]^0.5 / [NOBr(g)]

Given:
Initial NOBr concentration = 0.782 mol
Reaction volume = 100 L
Bromine concentration at equilibrium [Br2] = 1.42x10^-3 M

To find the equilibrium concentrations of NO and Br2, we need to use stoichiometry and the given information. Let's assume that x is the change in concentration for the decomposition of NOBr.

At equilibrium, the concentration of NOBr will be (0.782 - x) mol/L,
the concentration of NO will be x mol/L, and
the concentration of Br2 will be 0.5x mol/L.

Using the equation [Br2] = 0.5x, we can solve for x:

0.5x = 1.42x10^-3
x = 2 * (1.42x10^-3)
x = 2.84x10^-3 mol/L

Now, we can substitute these equilibrium concentrations into the equilibrium expression:

Kc = [NO(g)] * [Br2(g)]^0.5 / [NOBr(g)]
Kc = (2.84x10^-3) * [(2.84x10^-3 * 0.5)^0.5] / (0.782 - 2.84x10^-3)

Calculating this expression will give the value of Kc in M^0.5.

To calculate the equilibrium constant, Kc, for the given reaction, we need to find the concentrations of the reactants and products at equilibrium.

Let's start by analyzing the information given in the problem. We know that the initial amount or moles of NOBr is 0.782 mol, and the reaction occurs in a 100 L container. Therefore, the initial concentration of NOBr can be calculated as:

Initial concentration of NOBr = (moles of NOBr) / (volume of container)
= 0.782 mol / 100 L
= 0.00782 M

At equilibrium, we are given that the concentration of bromine, [Br2], is 1.42x10-3 M. However, we need to find the concentration of NO and Br2 in terms of M^0.5.

Since the stoichiometric coefficient for NO is 1 and for Br2 is 0.5, the concentration of NO can be assumed as [NO] = [Br2]^(1/2).

Therefore, [NO] = (1.42x10-3 M)^(1/2)
= 3.77x10-4 M^0.5

Now that we have the concentrations of all the species at equilibrium, we can substitute those values into the expression for Kc:

Kc = ([NO] * [Br2]^0.5) / [NOBr]

Kc = (3.77x10-4 M^0.5 * (1.42x10-3 M)^(1/2)) / (0.00782 M)

Thus, the equilibrium constant, Kc, for the given reaction at the specified conditions is the calculated value from the above equation.