Asample of nitrosyl bromide (nobr) decomposes according to the equation 2nobr(g)⇌2no(g)+br2(g) an equilibrium mixture in a 5.00-l vessel at 100 ∘c contains 3.26 g of nobr, 3.06 g of no, and 8.15 g of br2. part a calculate kc.

Kc = 0.122.

Explanation:

We can calculate Kc from the relation: Kc = [NO]².[Br₂] / [NOBr]².Firstly, we convert the number of grams of each component to moles using (n = mass / molar mass).n of NOBr = (3.26 g / 109.8 g/mole) = 0.03 moles.n of NO = (3.06 g / 29.9 g/mole) = 0.102 moles.n of Br₂ = (8.15 g / 159.9 g/mole) = 0.051 moles.Then we can get the concentration of each component by dividing its no. of moles by the volume of the vessel (concentration = n / V).The concentration of NOBr = (0.03 mole / 5.00 L) = 6x10⁻³ mol/L.The concentration of NO = (0.102 mole / 5.00 L) = 2x10⁻² mol/L.The concentration of Br₂ = (0.051 mole / 5.00 L) = 1x10⁻² mol/L.Now, we can get Kc:Kc = [NO]².[Br₂] / [NOBr]² = (2x10⁻² mol/L)².(1x10⁻² mol/L) / (6x10⁻³ mol/L)² = 0.122.

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