Question

1. consider the following reaction:

h₂(g) + i₂(g) ⇌ 2 hi(g). kc = 54.3 at 430 °c.
the initial concentrations of h₂, i₂, and hi are 0.00623 m, 0.00414 m, and 0.0224 m, respectively.
determine the concentrations of all three species at equilibrium at 430 °c.

Explanation:

Step1: Set up the ICE table

Let $x$ be the change in concentration of $H_2$ and $I_2$ at equilibrium.

	$H_2$	$I_2$	$HI$
Change (M)	-$x$	-$x$	+$2x$
Equilibrium (M)	0.00623 - $x$	0.00414 - $x$	0.0224+ $2x$

Step2: Write the equilibrium - constant expression

The equilibrium - constant expression for the reaction $H_2(g)+I_2(g)
ightleftharpoons2HI(g)$ is $K_c=\frac{[HI]^2}{[H_2][I_2]}$.
We know that $K_c = 54.3$, so $54.3=\frac{(0.0224 + 2x)^2}{(0.00623 - x)(0.00414 - x)}$.

Step3: Expand and simplify the equation

Expand the numerator: $(0.0224 + 2x)^2=0.0224^2+2\times0.0224\times2x+(2x)^2=0.00050176 + 0.0896x+4x^2$.
Expand the denominator: $(0.00623 - x)(0.00414 - x)=0.00623\times0.00414-0.00623x-0.00414x+x^2=0.0000257922-0.01037x + x^2$.
The equation becomes $54.3=\frac{0.00050176 + 0.0896x+4x^2}{0.0000257922-0.01037x + x^2}$.
Cross - multiply: $54.3(0.0000257922-0.01037x + x^2)=0.00050176 + 0.0896x+4x^2$.
$0.00140052646-0.562091x + 54.3x^2=0.00050176 + 0.0896x+4x^2$.
Rearrange to get a quadratic equation: $(54.3 - 4)x^2-(0.562091 + 0.0896)x+(0.00140052646 - 0.00050176)=0$.
$50.3x^2-0.651691x + 0.00089876646 = 0$.

Step4: Solve the quadratic equation

The quadratic formula for $ax^2+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 50.3$, $b=-0.651691$, and $c = 0.00089876646$.
First, calculate the discriminant $\Delta=b^2 - 4ac=(-0.651691)^2-4\times50.3\times0.00089876646$.
$\Delta = 0.424701-0.180477=0.244224$.
$x=\frac{0.651691\pm\sqrt{0.244224}}{2\times50.3}=\frac{0.651691\pm0.4942}{100.6}$.
We get two solutions for $x$. Taking the appropriate value (the one that makes physical sense in terms of concentrations, i.e., non - negative concentrations at equilibrium), $x = 0.0016$.

Step5: Calculate the equilibrium concentrations

$[H_2]=0.00623 - x=0.00623-0.0016 = 0.00463$ M.
$[I_2]=0.00414 - x=0.00414-0.0016 = 0.00254$ M.
$[HI]=0.0224+2x=0.0224 + 2\times0.0016=0.0256$ M.

Answer:

$[H_2]=0.00463$ M, $[I_2]=0.00254$ M, $[HI]=0.0256$ M