Question

Consider the reaction where glucose is oxidized to ethanol and carbon dioxide: C6H12O6(s) → 2 C2H5OH (l) + 2 CO2(g)

4.1 Calculate ΔrH°m (in kJ/mol ) at 25°C.
4.2 Calculate ΔrH°m (in kJ/mol ) at 60°C.

The answers are

ΔrHom(25°C)   =

-66.6

kj/mol

ΔrHom(60°C)=

-63.8

kj/mol

please show the work

Answer 1

C6H12O6(s) → 2 C2H5OH (l) + 2 CO2(g)

HRxn(25°C) = Hproducts(25°C) - HReactants(25°C)

HRxn = (2*C2H5OH + 2*CO2) - (C6H12O6)

HRxn = (2*-277.6+ 2*-393.9) - (-1273.3 ) = -69.7 kJ/mol (for exact values, substitute your data)

For H(60°C)

HRxn = Hrxn(25°C) + Hheating(25-60°)

Hating(25-60°) = Cpmix * (Tf-Ti) = Cpmix ( 60-25 )he

Cpmix(25-60°) = 2*Cp ethanol + 2 Cp CO2 - (1*Cp glucose)

Cpmix(25-60°) = 2*111.53 + 2 *36.61 - (1*218.6 ) = 77.68 J/K

Hating(25-60°) = Cpmix ( 60-25 ) = 77.68 * (60-25) = 2718.8 J = 2.719 kJ

HRxn = Hrxn(25°C) + Hheating(25-60°) = -69.7 + 2.719 = -66.9 kJ/mol

NOTE: use your specific data for 100% accuracy;