# For the reaction H2+I2----2HI Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?

Jun 25, 2017

$\left[\text{HI}\right] = 0.778 M$

#### Explanation:

We're asked to find the equilibrium concentration of $\text{HI}$ with an initial concentration of $1$ ${\text{mol H}}_{2}$ and $1$ ${\text{mol I}}_{2}$.

The equilibrium constant expression for this reaction is

${K}_{c} = \left(\left[{\text{HI"]^2)/(["H"_2]["I}}_{2}\right]\right) = 49$

The initial concentrations of both ${\text{H}}_{2}$ and ${\text{I}}_{2}$ are

$\left(1 \textcolor{w h i t e}{l} \text{mol")/(2color(white)(l)"L}\right) = 0.5 M$

So our initial concentrations for each species are

Initial:

• ${\text{H}}_{2}$: $0.5 M$

• ${\text{I}}_{2}$: $0.5 M$

• $\text{HI}$: $0$

From the coefficients of the chemical equation, we can predict the changes in concentration with the quantity $x$:

Change:

• ${\text{H}}_{2}$: $- x$

• ${\text{I}}_{2}$: $- x$

• $\text{HI}$: $+ 2 x$

and the final concentrations are the sum of the initial and change:

Final:

• ${\text{H}}_{2}$: $0.5 M - x$

• ${\text{I}}_{2}$: $0.5 M - x$

• $\text{HI}$: $2 x$

It's algebra time! Let's plug these into our equilibrium constant expression to start solving for $x$:

${K}_{c} = \frac{{\left(2 x\right)}^{2}}{\left(0.50 - x\right) \left(0.50 - x\right)} = 49$

$4 {x}^{2} = 49 \left(0.25 - x + {x}^{2}\right)$

$49 {x}^{2} - 49 x + 12.25 = 4 {x}^{2}$

$45 {x}^{2} - 49 x + 12.25 = 0$

$x = \frac{49 \pm \sqrt{{\left(- 49\right)}^{2} - 4 \left(45\right) \left(12.25\right)}}{2 \left(45\right)} = 0.7$ or $0.389$

We can neglect the solution that is greater than $0.5$, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is color(blue)(0.389.

Therefore, the final equilibrium concentration of $\text{HI}$ is

["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M