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 Δ_rH⁰ at T = 298 K.

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

ΔrH∘ =

kJ/mol

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Part B - Calculate Δ_rS⁰ at T = 298 K.

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

ΔrS∘ =

J/(mol.K)

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Part C - Calculate Δ_rG0 at T = 298 K.

Use FOUR significant figures.

ΔrG∘ =

kJ/mol

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Part D - Calculate ΔrH0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  ΔrHo310−ΔrHo298 )

Use three significant figures.

ΔrHo310−ΔrHo298 =

kJ/mol

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Part E - Calculate Δ_rS0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  ΔrSo310−ΔrSo298 )

Use three significant figures.

ΔrSo310−ΔrSo298 =

J/(mol.K)

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Part F - Calculate Δ_rG0 at T = 310 K.

Report the difference between this value and the value calculated at 298K (that is,  ΔrGo310−ΔrGo298 )

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

ΔrGo310−ΔrGo298 =

J/mol

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Answer 1

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 ΔrH0 at T = 298 K.

ΔrH0 = Σ ΔH°f (products) - Σ ΔH°f (reactants)

ΔrH0 = 2(– 673.6 kJ /mol ) – ( –1273.1 kJ/mol)

ΔrH0 = –74.10 kJ/mol

Part B - Calculate ΔrS0 at T = 298 K.

ΔrS0 = Σ ΔS° (products) - Σ ΔS° (reactants)

ΔrS0 = 2(192.1 J K–1mol) – (209.2 J K–1mol)

ΔrS0 = 175.02 J K–1mol

Part C - Calculate ΔrG0 at T = 298 K

ΔrG0 = ΔH - T ΔS

ΔrG0 = (–74.10 kJ/mol ) - 298(175.02/1000 k J K–1mol)

ΔrG0 = - 126.25 kJ/mol

Part D - Calculate ΔrH0 at T = 310 K.

To calculate DG° at T = 310 K

ΔH 310 K = ΔH 298 K + ΔCp ΔT

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

ΔH 310 K = - 76.5 kJ/mol

ΔrHo310−ΔrHo298 = (- 76.5 kJ/mol) –(-74.10 kJ/mol)

ΔrHo310−ΔrHo298 = -2.40 kJ/mol (three significant figures)