Question

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3OH] = 1.56 M.

Answer 1

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

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

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

• Kc= 14.5
• [H₂]= 0.322 M
• [CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M