3. (3 points) (Adapted from Atkins) Borneol (C10H1sO is a terpene occurring in c

Question

3. (3 points) (Adapted from Atkins) Borneol (C10H1sO is a terpene occurring in certain essential oils derived from plants. It is used in traditional medicine, and also can be used as an insect repellant. When its single OH group is in the axial position it is called borneol (B), with the OH in the equatorial position it is called isoborneol (I). You will know the molecule when the OH is oxidized to a ketone (-O): camphor commonly used in moth balls. Consider the gas phase equilibrium between borneol and isoborneol at 503K: G°(T = 503 K) = +9.4 kJ Calculate the final concentrations (mol/L) of borneol (B) and isoborneol (I) when initially 0.80 g borneol and 0.40 g isoborneol are introduced in a 250 mL closed container and the temperature is raised to 503K.

Explanation / Answer

The free energy change, G0(T = 503 K) is given. Use the supplied information to find out the value of the equilibrium constant, K at 503 K.

We have G0 = -R*T*ln K where R is the gas constant and T is the Kelvin temperature. Plug in values and get

(+9.4 kJ) = -(8.314 J/mol.K)*(503 K)*ln K

=====> (+9.4 kJ)*(1000 J/1 kJ) = -(8.314 J/mol.K)*(503 K)*ln K

=====> ln K = -2.2477

=====> K = exp^(-2.2477) = 0.1056

The equilibrium constant for the reaction at 503 K is calculated to be 0.1056.

Molar mass of borneol, B = molar mass of isoborneol, I = 154.25 g/mol; mole(s) of B corresponding to 0.80 g = (0.80 g)/(154.25 g/mol) = 0.005186 mole; mole(s) of I corresponding to 0.40 g = (0.40 g)/(154.25 g/mol) = 0.002593 mole.

Next, set up the ICE chart for the reaction (in terms of number of moles reacted) as below.

B (g) <========> I (g)

initial                                             0.005186                  0.002593

change                                              -x                             +x

equilibrium                              (0.005186 – x)            (0.002593 + x)

[B] = moles of B at equilibrium/volume of container in L = (0.005186 – x)/[(250 mL)*(1 L/1000 mL)] = (0.005186 – x)/0.25 mol/L

[I] = moles of I at equilibrium/volume of container in L = (0.002593 + x)/[(250 mL)*(1 L/1000 mL)] = (0.002593 + x)/0.25 mol/L

Write down the expression for K as below.

K = [I]/[B]

====> 0.1056 = [(0.002593 + x)/0.25 mol/L]/[(0.005186 – x)/0.25 mol/L]

====> 0.1056 = (0.002593 – x)/(0.005186 + x)

====> 0.1056*(0.005186 – x) = 0.002593 + x

====> 0.00054764 – 0.1056x = 0.002593 + x

====> x + 0.1056x = 0.002593 – 0.00054764

====> 1.1056x = 0.00204536

====> x = 0.00204536/1.1056 = 0.001850

The molar concentrations are [B] = (0.005186 – 0.001850)/0.25 mol/L = 0.013344 mol/L (ans)

and [I] = (0.002593 + 0.001850)/0.25 mol/L = 0.017772 mol/L (ans).

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