Question

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to reach equilibrium. Calculate the concentration of HI at equilibrium.

Answer 1

Answer:

The concentration of HI at equilibrium is 0.2445 $\frac{moles}{L}$

Explanation:

A chemical reaction occurs in both directions: from reagents transforming into products (direct reaction) and from products transforming back into reactants (reverse reaction)

The mathematical expression that  represents the Chemical Balance is the equilibrium constant Kc.

You have:

aA + bB ⇔ cC + dD

where A, B, C and D represent the chemical species involved and a, b, c and d their respective stoichiometric coefficients. So the constant Kc is:

$Kc=\frac{[A]^{a}*[B]^{b} }{[C]^{c}*[D]^{d} }$

That is, this constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also elevated to their stoichiometric coefficients.

So, in this case:

$Kc=\frac{[H_{2}]*[I_{2} ] }{[HI]^{2} }=0.0156$

You have that the initial concentrations are:

[HI]= $\frac{0.550 moles}{2.00 L} = 0.275 \frac{moles}{L}$

[H₂]= 0

[I₂]= 0

Being "x" the change in the concentration that occurs during the reaction, which must be affected by the stoichiometric coefficient, the final concentrations of each species in equilibrium will be:

[HI]= $0.275 \frac{moles}{L}-x$

[H₂]= x

[I₂]= x

Keep in mind that in the case of reagents the concentration "x" is subtracted because the reagents are consumed. In the case of products, the concentration "x" is added because the reagents are formed.

Then:

$0.0156=\frac{x*x}{(0.275-x)^{2} }$

Resolving

0.0156*(0.275-x)²=x²

0.0156*(0.075625-0.55*x+x²)=x²

1.17975*10⁻³-8.58*10⁻³*x+0.0156*x²=x²

-0.9844*x²-8.58*10⁻³*x+1.17975*10⁻³=0

Solving for  x  will get you two values: x1≅0.0305 and x2≅-0.0392

Since the value of "x" represents a concentration, and cannot have negative values, the value of x2 is discarded. So: x=x1

Then:

[HI]= $0.275 \frac{moles}{L}-x$=$0.275 \frac{moles}{L}-0.0305 \frac{moles}{L} = 0.2445 \frac{moles}{L}$

[H₂]= 0.0305 $\frac{moles}{L}$

[I₂]= 0.0305 $\frac{moles}{L}$

The concentration of HI at equilibrium is 0.2445 $\frac{moles}{L}$

Answer 2

Answer:

0.220

Explanation: