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

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

Answer 2

Answer:

(a) $p_{C6H5CH2OH}=2.14x10^{-3}atm$

(b) $Dissociation =0.99$

Explanation:

Hello,

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

$Kc=\frac{[C6H5CHO][H2]}{[C6H5CH2OH]}$

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

$[C6H5CH2OH]_0=\frac{1.20g*\frac{1mol}{108.14g} }{2.00L} =5.55x10^{-3}M$

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

$Kc=\frac{(x)(x)}{5.55x10^{-3}-x}=0.558$

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:

$p_{C6H5CH2OH}=[C6H5CH2OH]_{eq}RT =5x10^{-5}\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*523K \\p_{C6H5CH2OH}=2.14x10^{-3}atm$

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

$Dissociation=\frac{C6H5CH2OH_{eq}}{C6H5CH2OH_0} =\frac{5.50x10^{-3}M}{5.55x10^{-3}M} =0.99$

Best regards.