Question

Be sure to answer all parts.The equilibrium constant Kc for the reaction below is 61.5 at a certain temperature. H2(g) + I2(g) ⇆ 2HI(g) If you start with 0.155 M hydrogen iodide, what will the concentrations of H2, I2, and HI be at equilibrium?

Answer 1

Answer:

$[I_2]=[H_2]=0.0157M$

$[HI]=0.124M$

Explanation:

Hello!

In this case, for the undergoing chemical reaction, we can write the equilibrium expression as shown below:

$Kc=\frac{[HI]^2}{[I_2][H_2]}$

However, since the reactions starts with 0.155 M hydrogen iodide, we need to invert the reaction as well as the equilibrium expression:

$2HI\rightarrow I_2+H_2$

$\frac{1}{Kc}=0.0163 =\frac{[I_2][H_2]}{[HI]^2}$

And each concentration at equilibrium in terms of the reaction extent is:

$[I_2]=[H_2]=x$

$[HI]=0.155-2x$

Thus, plugging in the new expression we write:

$0.0163 =\frac{x*x}{(0.155-2x)^2}$

Which has the following valid solution for $x$:

$x=0.0157M$

Because the other solution is negative. Therefore, the concentration of hydrogen, iodine and hydrogen iodide at equilibrium turns out:

$[I_2]=[H_2]=0.0157M$

$[HI]=0.155-2(0.0157)=0.124M$

Best regards!

Answer 2

At the equilibrium, the concentration of hydrogen and iodine has been 0.0157 M each, and the concentration of HI will be 0.124 M.

The initial concentration of the reactant has been 0.155 M. From the balanced chemical equation, the ICE table will be:

         HI          $\rm H_2$          $\rm I_2$

I      0.155        0           0

C       -2x         x           x

E  0.155 - 2x    x          x

The value of the equilibrium constant can be given by the ratio of the concentration of products to the reactant.

The equilibrium constant for the reaction will be:

$\rm k_c\;=\;\dfrac{[H_2]\;[I_2]}{[HI]^2}$

The given constant value has been for the opposite reaction. Thus the constant for the required reaction has been :

$\rm \dfrac{1}{given\;k_c} \;=\;\dfrac{[H_2]\;[I_2]}{[HI]^2}$

$\rm \dfrac{1}{61.5}$ = $\rm \dfrac{(x)\;(x)}{(0.155-2x)^2}$

0.0162 = $\rm \dfrac{x^2}{(0.155-2x)^2}$

By simplyfying the equation,

x = 0.0157 M

At the equilibrium, the concentration of hydrogen and iodine has been 0.0157 M each.

The concentration of HI will be:

HI = 0.155 - 2x

HI = 0.155 - 2(0.0157)

HI = 0.124 M.          

For more information about the equilibrium constant, refer to the link:

https://brainly.com/question/10038290