Question

Enter your answer in the provided box.The equilibrium constant Kc for the equation2H2(g) + CO(g) ⇌ CH3OH(g)is 35 at a certain temperature. If there are 3.21 ×10−2 moles of H2 and 4.87 ×10−3 moles of CH3OH at equilibrium in a 3.63−L flask, what is the concentration of CO?

Answer 1

Answer:

The concentration of CO is 4.33*10⁻³$\frac{moles}{L}$

Explanation:

A reversible chemical reaction, as in this case, occurs in both directions: reagents transforming into products (direct reaction) and products transforming back into reagents (reverse reaction).

Chemical Equilibrium is the state in which the direct and indirect reaction have the same reaction rate and is defined by a chemical constant Kc.

Being:

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, constant Kc is defined as:

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

In the case of the reaction

2 H₂(g) + CO(g) ⇌ CH₃OH(g)

the equilibrium constant Kc is:

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

You know:

• Kc=35

• [CH₃OH] =$\frac{4.87*10^{-3}moles }{3.63 L} =1.34*10^{-3} \frac{moles}{L}$

• [H₂]=$\frac{3.21*10^{-2}moles }{3.63 L} =8.84*10^{-3} \frac{moles}{L}$

• [CO]=?

Replacing:

$35=\frac{1.34*10^{-3} }{(8.84*10^{-3} )^{2} *[CO]}$

Solving you get:

$[CO]= \frac{1.34*10^{-3} }{(8.84*10^{-3} )^{2} *35}$

[CO]=4.33*10⁻³$\frac{moles}{L}$

The concentration of CO is 4.33*10⁻³$\frac{moles}{L}$