Question

For the reaction i2(g)+br2(g)←−→2ibr(g), kc=280 at 150 ∘c. suppose that 0.520 mol ibr in a 2.00-l flask is allowed to reach equilibrium at 150 ∘c. part a what is the equilibrium concentration of ibr?

Answer 1

first, we have to get the initial [IBr] = 0.520 mol / 2 L =0.26 M
According to the reaction balanced equation:
and using ICE table :
             I2(g) + Br2(g)↔ 2IBr(g)
initial     0           0            0.26
Change +X        +X            -2X
Equ          X          X          (0.26-2X)

when Kc = [IBr]^2/[I2][Br2]

so by substitution:
280 = (0.26-2X)^2 / X^2 by solving this equation for X
∴X = 0.0139 
∴[I2] = [Br]2 = 0.0139
and [IBr] = 0.26- (2*0.0139)
              = 0.2322 M

Answer 2

The equilibrium concentration of IBr is $\boxed{{\text{0}}{\text{.23226 M}}}$.

Further Explanation:

Chemical equilibrium

It is a condition at which the rate of forward reaction is equal to that of backward reaction. At equilibrium, the formation of product from reactant gets balanced out by the formation of reactants from products so there is no change in concentrations of both reactants and products.

Equilibrium can be generally represented by the following equation:

 $a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}$

Here,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The formula to calculate the equilibrium constant for the general reaction is as follows:

 ${K_{\text{c}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$

Here,

${K_{\text{c}}}$ is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

The given reaction occurs as follows:

 $2{\text{IBr}}\left( g \right) \rightleftharpoons {{\text{I}}_2}\left( g \right) + {\text{B}}{{\text{r}}_{\text{2}}}\left( g \right)$

The concentration of IBr can be calculated as follows:

 $\begin{aligned}\left[ {{\text{IBr}}} \right]&=\frac{{0.520{\text{ mol}}}}{{2{\text{ L}}}}\\&=0.260{\text{ M}}\\\end{aligned}$

The initial concentration of IBr is 0.260 M. Let x to be the change in concentration at equilibrium. Therefore, the concentration of IBr becomes 0.260-2x at equilibrium. The concentration of both ${{\text{I}}_{\text{2}}}$ and ${\text{B}}{{\text{r}}_{\text{2}}}$ become x at equilibrium.

The expression of ${K_{\text{c}}}$ for the above reaction is as follows:

${K_{\text{c}}} = \dfrac{{\left[ {{{\text{I}}_2}} \right]\left[ {{\text{B}}{{\text{r}}_2}} \right]}}{{{{\left[ {{\text{IBr}}} \right]}^2}}}$   …… (1)                                                                            

The value of equilibrium constant $\left( {K_{\text{c}}^'} \right)$ for the above reaction is calculated as follows:

$\begin{aligned}K_{\text{c}}^' &= \frac{1}{{{\text{280}}}}\\&= 0.00357\\\end{aligned}$  

Substitute x for ${{\text{I}}_{\text{2}}}$, x for ${\text{B}}{{\text{r}}_{\text{2}}}$, 0.260-2x for [IBr] and 0.00357 for ${K_{\text{c}}}$ in equation (1).

$0.00357 = \dfrac{{\left( {\text{x}} \right)\left( {\text{x}} \right)}}{{{{\left( {0.260 - 2x} \right)}^2}}}$           …… (2)                                            

Rearrange equation (2) to obtain the quadratic equation as follows:

 ${x^2} + 0.003767x + 0.0002448 = 0$

Solve for x,

${\text{x}} =  - 0.01765$  

Or,

${\text{x}} = 0.01387$

The value of ${\text{x}} =  - 0.01765$ is rejected as concentration cannot be negative. So the value of x is 0.01387.

The concentration of IBr at equilibrium is calculated as follows:

 $\begin{aligned}\left[ {{\text{IBr}}} \right] &= 0.260 - 2\left( {0.01387} \right)\\&= {\text{ 0}}{\text{.23226 M}}\\\end{aligned}$

Therefore the equilibrium concentration of IBr is 0.23226 M.

Learn more:

1. Calculation of equilibrium constant of pure water at ${\text{25 }}^\circ {\text{C}}$ : https://brainly.com/question/3467841
2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: chemical equilibrium, reactants, products, concentration, A, B, C, D, a, b, c, d, kc, equilibrium constant, 0.23226 M.