Under anaerobic conditions, glucose is broken down in muscle tissue to form lact

Question

Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction:

C6H12O6 2 CH3CHOHCOOH

Use data from the 'official' tables available in the BCH341 website in addition to the heat capacities for glucose and lactic acid given below:

Assume all heat capacities are constant from T = 298 K to T = 310 K.

For glucose,  Cop,m=218.2JK.mol

For lactic acid,  Cop,m=127.6JK.mol

Note: All thermodynamic quantities in this problem are per mol of glucose.

Be sure you pay attention to units and significant figures. Rounding errors can lead to significant deviations from the correct answers, so be careful!

Part A - Calculate rH0 at T = 298 K.

Express your answer in kJ/mol to FOUR significant figures.

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Part B - Calculate rS0 at T = 298 K.

Express your answer in J/(K.mol) to FOUR significant figures.

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Part C - Calculate rG0 at T = 298 K.

Use FOUR significant figures.

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Part D - Calculate rH0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  rHo310rHo298 )

Use three significant figures.

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Part E - Calculate rS0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  rSo310rSo298 )

Use three significant figures.

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Part F - Calculate rG0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  rGo310rGo298 )

Use three significant figures. Express your answer in J/mol.

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Explanation / Answer

Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction:

C6H12O6 2 CH3CHOHCOOH

Assume all heat capacities are constant from T = 298 K to T = 310 K.

For glucose DHf ° –1273.1 kJ mol -1

Cp,m =218.2JK.mol

S° =209.2 J K–1mol

For lactic acid,

DHf ° = – 673.6 kJ mol -1

Cp,m = 127.6JK.mol

S° = 192.1 J K–1mol

Part A - Calculate rH0 at T = 298 K.

rH0 = H°f (products) - H°f (reactants)

rH0 = 2(– 673.6 kJ /mol ) – ( –1273.1 kJ/mol)

rH0 = –74.10 kJ/mol

Part B - Calculate rS0 at T = 298 K.

rS0 = S° (products) - S° (reactants)

rS0 = 2(192.1 J K–1mol) – (209.2 J K–1mol)

rS0 = 175.02 J K–1mol

Part C - Calculate rG0 at T = 298 K

rG0 = H - T S

rG0 = (–74.10 kJ/mol ) - 298(175.02/1000 k J K–1mol)

rG0 = - 126.25 kJ/mol

Part D - Calculate rH0 at T = 310 K.

To calculate DG° at T = 310 K

H 310 K = H 298 K + Cp T

H 310 K =–74.10 kJ/mol + [ 2 x (127.6/1000kJK.mol )+ -1(218.2/1000kJK.mol) x 12K]

H 310 K = - 76.5 kJ/mol

rHo310rHo298 = (- 76.5 kJ/mol) –(-74.10 kJ/mol)

rHo310rHo298 = -2.40 kJ/mol (three significant figures)

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