Question

The dehydrogenation of benzyl alcohol to make the flavoring agent benzaldehyde is an equilibrium process described by the equation: C6H5CH2OH(g) ⇆ C6H5CHO(g) + H2(g) At 523 K, the value of its equilibrium constant is K = 0.558. (a) Suppose that 1.20 g of benzyl alcohol is placed into a 2.00 L vessel and heated to 523 K. What is the partial pressure of benzaldehyde when equilibrium is attained? (b) What fraction of benzyl alcohol is dissociated into products at equilibrium?

Answer 1

pC6H5CHO = 0.180 atm

Fraction dissociated = 0.756

Step-by-step 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

Answer 2

(a)
`p_(C6H5CH2OH)=2.14x10^(-3)atm`

(b)
`Dissociation =0.99`

Step-by-step explanation:

Hello,

(a) In this case, for the given chemical reaction, the law of mass action becomes:

`Kc=([C6H5CHO][H2])/([C6H5CH2OH])`

In such a way, as 1.20 g of benzyl alcohol are placed into a 2.00-L vessel, the initial concentration is:

`[C6H5CH2OH]_0=(1.20g*(1mol)/(108.14g) )/(2.00L) =5.55x10^(-3)M`

Hence, by writing the law of mass action in terms of the change
`x` due to equilibrium:

`Kc=((x)(x))/(5.55x10^(-3)-x)=0.558`

Solving for
`x` by using a quadratic equation one obtains:

`x=5.50x10^(-3)M`

Thus, the equilibrium concentration of benzyl alcohol is computed:

`[C6H5CH2OH]_(eq)=5.55x10^(-3)M-5.50x10^(-3)M=5x10^(-5)M`

With that concentration the partial pressure results:

`p_(C6H5CH2OH)=[C6H5CH2OH]_(eq)RT =5x10^(-5)(mol)/(L) *0.082(atm*L)/(mol*K)*523K \\p_(C6H5CH2OH)=2.14x10^(-3)atm`

(b) Now, the fraction of benzyl alcohol that is dissociated relates its equilibrium concentration with its initial concentration:

`Dissociation=(C6H5CH2OH_(eq))/(C6H5CH2OH_0) =(5.50x10^(-3)M)/(5.55x10^(-3)M) =0.99`

Best regards.