Respuesta :
Answer:
For a: The equilibrium concentration of CO and [tex]H_2[/tex] are 0.24 M and 0.32 M.
For b: The value of [tex]K_c\text{ and }K_p[/tex] are 1.5625 and [tex]8.41\times 10^{-4}[/tex]
Explanation:
We are given:
Volume of container = 5 L
Initial moles of CO = 1.8 moles
Initial concentration of CO = [tex]\frac{1.8mol}{5L}[/tex]
Initial moles of [tex]H_2[/tex] = 2.2 moles
Initial concentration of [tex]H_2[/tex] = [tex]\frac{2.2mol}{5L}[/tex]
Equilibrium moles of [tex]CH_3OH[/tex] = 0.6
Equilibrium concentration of [tex]CH_3OH[/tex] = [tex]\frac{0.6mol}{5L}=0.12M[/tex]
- For a:
The chemical equation for the formation of methanol follows:
[tex]CO(g)+H_2(g)\rightleftharpoons CH_3OH(l)[/tex]
t = 0 [tex]\frac{1.8}{5}[/tex] [tex]\frac{2.2}{5}[/tex] 0
[tex]t=t_{eq}[/tex] [tex]\frac{1.2}{5}[/tex] [tex]\frac{1.6}{5}[/tex] [tex]\frac{0.6}{5}[/tex]
So, the equilibrium concentration of CO = [tex]\frac{1.2}{5}=0.24M[/tex]
The equilibrium concentration of [tex]H_2[/tex] = [tex]\frac{1.6}{5}=0.32M[/tex]
- For b:
The expression of [tex]K_c[/tex] for the given chemical reaction follows:
[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]}[/tex]
We are given:
[tex][CH_3OH]=0.12mol/L[/tex]
[tex][CO]=0.24mol/L[/tex]
[tex][H_2]=0.32mol/L[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{0.12}{0.24\times 0.32}=1.5625[/tex]
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
Where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?
[tex]K_c[/tex] = equilibrium constant in terms of concentration = 1.5625
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = 525 K
[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=0-2=-2[/tex]
Putting values in above equation, we get:
[tex]K_p=1.5625\times (0.0821\times 525)^{-2}\\\\K_p=8.41\times 10^{-4}[/tex]
Hence, the value of [tex]K_c\text{ and }K_p[/tex] are 1.5625 and [tex]8.41\times 10^{-4}[/tex]
