Question

Table sugar (sucrose) is hydrolyzed to form glucose (a simple sugar) according to the following equation: C12H220 i 1 (s)-H2O(g)→2C6H 1206(s) Determine the enthalpy change for the hydrolysis reaction given the following information: C12H22011 (s) + 1202(g) → 12CO2(g) + i i H2O(g) AH-5640 kJ C,H,..(s) + 602(g)-→ 6CO2(g) + 6H20(g) 6 126 H2800 kJ rxn kj

Answer 1

(1) C12H22O11(s) + 12O2(g) $\rightarrow$ 12CO2(g) + 11H2O(g)        $\Delta$H_rxn = -5640 kJ

(2) C2H12O6(s) + 6O2(g) $\rightarrow$ 6CO2(g) + 6H2O(g)                 $\Delta$H_rxn = -2800 kJ

Reversing the equation (2), we get

(3) 6CO2(g) + 6H2O(g) $\rightarrow$ C2H12O6(s) + 6O2(g)                 $\Delta$H_rxn = +2800 kJ

Multiplying the equation (2) by 2, we get

(4) 12CO2(g) + 12H2O(g) $\rightarrow$ 2C2H12O6(s) + 12O2(g)         $\Delta$H_rxn = 2 (+2800 kJ)
                                                                                                            = 5600 kJ

Adding equation (1) and (4), we get

C12H22O11(s) + 12O2(g) + 12CO2(g) + 12H2O(g) $\rightarrow$12CO2(g) + 11H2O(g) + 2C2H12O6(s) + 12O2(g)
   $\Delta$H_rxn = - 5640 kJ + 5600 kJ

C12H22O11(s) + H2O(g) $\rightarrow$ 2C2H12O6(s)   $\Delta$H_rxn = -40 kJ