Question

Knowing that the equilibrium constant for the reaction below H2 (g) + Br2 (g) ⇌ 2 HBr (g) is Kc = 3.8 × 104, determine the equilibrium constant for the following reaction: 4 HBr (g) ⇌ 2 H2 (g) + 2 Br2 (g)

Answer 1

Answer:

Kc new is the inverse of Kc, then $Kc_{new} = 2.63 x 10^{-5}$

Explanation:

We know the equilibium constant for this reaction:

$H_{2}+ Br_{2} <--> 2HBr$      

The expression of Kc is:

$Kc = \frac{[Products]^{x} }{[Reagents]^{y} }$  

for our reaction:

*Product is :   HBr

*Reagents are: H2, Br2

*x: coeficient of products (taking from the balanced chemical reaction)

*y: coefficient of each reagent (taking from the balanced chemical reaction)

Considering these information,  expression of Kc is:

$Kc = \frac{[HBr]^{2} }{[H_{2}] * [Br_{2}] }$ and the value of Kc is Kc = $3.8 x 10^{4}$

The question of this exercise is to determine Kc for the same reaction, but the inverse one, now Reactive is HBr and Products are H2 and Br2.  Coefficients are differents because they are multiple of 2.

Balanced reaction is:

$4HBr <--> 2H_{2} + 2Br_{2}$

simplifiying coefficients of the balanced chemical reaction, new reaction is:

$2 HBr <--> H_{2} + Br_{2}$,    

Expression of $Kc_{new}$ is:

$Kc_{new} = \frac{H_{2}*Br_{2} }{[HBr]^{2} }$

If we compare expression of Kc and with expression of $Kc_{new}$, we can see  $Kc_{new} = \frac{1}{Kc}$

Finally, $Kc_{new} =\frac{1}{3.8x10^{4} } = 2.63 x 10^{-5}$