Question

At $800{\text{ }}K$ hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is $5 \times {10^8}$. Calculate the amount of ${H_2}, B{r_2} and {\text{ }}HBr$ at equilibrium if a mixture of $0.6{\text{ }}mol$ of ${H_2}$ and $0.2{\text{ }}mol$ of bromine is heated to $700{\text{ }}K$.

Answer 1

First we have to develop the expected equation and use it to find the value of x, using that to find the final value of the compounds.
Formula used:
K = $\dfrac{{{{[HBr]}^2}}}{{[{H_2}][B{r_2}]}}$

Complete step by step answer:
Let’s start with formulating the reaction equation between the hydrogen $(H_2)$, Bromine $(Br_2)$ and their product $HBr$ at equilibrium
${H_2}(g){\text{ }} + {\text{ }}B{r_2}(g){\text{ }}\overset {} \leftrightarrows {\text{ }}2HBr$
The initial concentration of ${H_2}$ is 0.6mol and for $B{r_2}$ is 0.2mol. The equilibrium constant k of the reaction is $5{\text{ }} \times {\text{ }}{10^8}$. And the final after equilibrium will be $\left( {0.6 - x} \right)$ for $H_2$, 0.2-x for $Br_2$ and 2x for HBr.
We also know that K = $\dfrac{{{{[HBr]}^2}}}{{[{H_2}][B{r_2}]}}$, now replacing the values we get
K=$\dfrac{{4{x^2}}}{{(0.6 - x)(0.2 - x)}}$ which will be 5 X 108.
Solving for x we get x=0.6 or 0.2 from this $HBr$ is equal to 0.4 mol
Now, since it’s an Equilibrium reaction the reverse is also true which means
${\text{ 2HBr}}\overset {} \leftrightarrows {\text{ }}{{\text{H}}_{\text{2}}}{\text{(g) + B}}{{\text{r}}_{\text{2}}}{\text{(g)}}$

At equilibrium	$\left( {0.4 - 2x} \right)$	$\left( {0.4 + x} \right)$	$x$

$K{\text{ }} = {\text{ }}\left( {0.4{\text{ }} + {\text{ }}x} \right)x/{\left( {0.4 - 2x} \right)^2}$
$x{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}{10^{ - 10}}mole$
So, $HBr$ will be equal to 0.4 mole, $B{r_2}$ will be $2{\text{ }} \times {\text{ }}{10^{ - 10}}$ mole and ${H_2}$ will be 0.4 mole.

Note:
We must know that the reaction equilibrium is the realistic case which tells that the reaction is proceeding in both the direction, forward and backward direction. The direction of the reaction depends on the amount of reactant we are adding and the product being formed. At equilibrium, the amount of product being formed is equal to the amount of reactant formed from the product.