Question

Hydrogen gas and bromine vapor react to form gaseous hydrogen bromide at 1300 K. The value of the equilibrium constant is 1.6 x 105. What is the value of the equilbrium constant for the decomposition of gaseous hydrogen bromide to form hydrogen gas and bromine vapor at 1300 K?

Answer 1

Often, the products of a chemical reaction can be combined with each other to form reagents again. The reactions in which this occurs are called reversible reactions. In these cases, the transformation of reactants into products is partial, reaching a state of chemical equilibrium in which the speeds of the direct reactions inverse are equalized.

The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium. The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

In this way, we must first write the reaction in question (do not forget to balance the reaction),

H₂ + Br₂ → 2HBr

For this reaction the equilibrium constant is

Kc = $\frac{[HBr]^{2} }{[H_{2} ][Br_{2} ]}$ = 1.6ₓ10⁵ at 1300 K.

The decomposition reaction of gaseous hydrogen bromide to form hydrogen gas and bromine vapor is

2HBr → H₂ + Br₂

Then, the equilibrium constant for this reaction is,

Kc' = $\frac{[H_{2} ][Br_{2} ] }{[HBr]^{2}}$

As you can notice this equilibrium constant is the inverse of Kc, so

Kc' = 1 / Kc = 1 / (1.6ₓ10⁵) → Kc' = 6.25ₓ10⁻⁶

So, the equilibrium constant for the decomposition of gaseous hydrogen bromide to form hydrogen gas and bromine vapor is 6.25ₓ10⁻⁶ at 1300 K.