Hydrogen is the lightest of elements with atomic number one. When hydrogen gas reacts with bromine gas, they form hydrogen bromide gas, i.e., H2(g) + Br2(g) ⇔ 2HBr(g)

Question

At equilibrium, concentration values of the compounds are:

Compound [H2(g)] [Br2(g)] [HBr(g)]
Equilibrium Concentration 0.024 mol/L 0.024 mol/L 0.05 mol/L

The value of equilibrium constant Keq, for the above reaction, is

a) 2.6
b) 3.1
c) 4.3
d) 5.2
e) 8.7

Answer 1

To calculate the value of the equilibrium constant Keq, we can use the formula:

Keq = [HBr(g)]^2 / ([H2(g)] * [Br2(g)])

Plugging in the equilibrium concentration values:

Keq = (0.05 mol/L)^2 / ((0.024 mol/L) * (0.024 mol/L))

Keq = 0.0025 mol^2/L^2 / 0.000576 mol^2/L^2

Keq ≈ 4.3403

The value of Keq, rounded to one significant figure, is approximately 4.3.

Therefore, the answer is c) 4.3.

Answer 2

are you sure?

Answer 3

I apologize for the mistake in my previous response. Let's recalculate.

The equilibrium constant (Keq) is calculated by taking the ratio of the molar concentrations of the products to the molar concentrations of the reactants, each raised to their respective stoichiometric coefficients:

Keq = ([HBr(g)]^2) / ([H2(g)] * [Br2(g)])

Plugging in the given equilibrium concentrations:

Keq = (0.05 mol/L)^2 / ((0.024 mol/L) * (0.024 mol/L))

Keq ≈ 0.0025 mol^2/L^2 / 0.000576 mol^2/L^2

Keq ≈ 4.3402

Therefore, the correct answer is c) 4.3.