Respuesta :
Answer:
(a) [tex]p_{C6H5CH2OH}=2.14x10^{-3}atm[/tex]
(b) [tex]Dissociation =0.99[/tex]
Explanation:
Hello,
(a) In this case, for the given chemical reaction, the law of mass action becomes:
[tex]Kc=\frac{[C6H5CHO][H2]}{[C6H5CH2OH]}[/tex]
In such a way, as 1.20 g of benzyl alcohol are placed into a 2.00-L vessel, the initial concentration is:
[tex][C6H5CH2OH]_0=\frac{1.20g*\frac{1mol}{108.14g} }{2.00L} =5.55x10^{-3}M[/tex]
Hence, by writing the law of mass action in terms of the change [tex]x[/tex] due to equilibrium:
[tex]Kc=\frac{(x)(x)}{5.55x10^{-3}-x}=0.558[/tex]
Solving for [tex]x[/tex] by using a quadratic equation one obtains:
[tex]x=5.50x10^{-3}M[/tex]
Thus, the equilibrium concentration of benzyl alcohol is computed:
[tex][C6H5CH2OH]_{eq}=5.55x10^{-3}M-5.50x10^{-3}M=5x10^{-5}M[/tex]
With that concentration the partial pressure results:
[tex]p_{C6H5CH2OH}=[C6H5CH2OH]_{eq}RT =5x10^{-5}\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*523K \\p_{C6H5CH2OH}=2.14x10^{-3}atm[/tex]
(b) Now, the fraction of benzyl alcohol that is dissociated relates its equilibrium concentration with its initial concentration:
[tex]Dissociation=\frac{C6H5CH2OH_{eq}}{C6H5CH2OH_0} =\frac{5.50x10^{-3}M}{5.55x10^{-3}M} =0.99[/tex]
Best regards.
Answer:
pC6H5CHO = 0.180 atm
Fraction dissociated = 0.756
Explanation:
Step 1: Data given
Temperature = 523 K
the value of its equilibrium constant is K = 0.558
Mass of benzyl alcohol = 1.20 grams
Molar mass of benzyl alcohol = 108.14 g/mol
Volume = 2.00 L
heated to 523 K
Step 2: The balanced equation
C6H5CH2OH(g) ⇆ C6H5CHO(g) + H2(g)
Step 3: Calculate moles benzyl alcohol
Moles benzyl alcohol = Mass / molar mass
Moles benzyl alcohol = 1.20 grams / 108.14 g/mol
Moles benzyl alcC6H5CH2OHohol = 0.0111 moles
Step 4: Initial moles
Moles C6H5CH2OH = 0.0111 moles
Moles C6H5CHO = 0 moles
Moles H2O = 0 moles
Step 5: moles at the equilibrium
Moles C6H5CH2OH = 0.0111 - X moles
Moles C6H5CHO = X moles
Moles H2O = X moles
Step 6: Calculate the total number of moles at equilibrium
Total number of moles = (0.0111 - X moles) + X moles + X moles
Total number of moles = 0.0111 + X moles
Step 7: Calculate the total pressure at the equilibrium
p*V = n*R*T
p = (n*R*T) / V
⇒with p = the total pressure at the equilibrium = TO BE DETERMINED
⇒with n = the total number of moles = 0.0111 + X moles
⇒with R = the gas constant = 0.08206 L*atm / mol * K
⇒with T = the temperature = 523 K
⇒with V = the volume of the vessel = 2.00 L
p = (0.0111 - X moles ) * 0.08206*523 / 2.00
p = 21.46(0.0111 - X moles)
Step 8: Define the equilibrium constant K
K = 0.558 = (pC6H5CHO)*(pH2) / (pC6H5CH2OH)
0.558 = (X / (0.0111 + X)*P)² / ((0.0111-X)/(0.0111+X)*P)
0.558 = (X²(21.46 * (0.0111+X))) / ((0.0111 + X) (0.0111-X))
X = 0.00839
Step 9: Calculate the equilibrium partial pressure
pC6H5CHO = X / (0.0111 + X) * (21.46 * (0.0111 +X))
pC6H5CHO = 0.180 atm
Step 10: What fraction of benzyl alcohol is dissociated into products at equilibrium?
Fraction dissociated = Δn / n°
Fraction dissociated = X / 0.0111
Fraction dissociated = 0.00839 / 0.0111
Fraction dissociated = 0.756
