Question

The ΔH of reaction for the combustion of glucose (C6H12O6) and of sacharose (C12H22O11) are respectively -2805 kJ and -5644kJ. Calculate the enthalpy for the reaction of hydrolysis of saccharose into two molecules of glucose:

C₁₂H₂₂O₁₁(s) + H₂O(l) --> 2C₆H₁₂O₆(s)

Answer 1

C6H12O6 + 6 O2 -------------------> 6CO2 + 6H2O , DH = -2805 kJ ------------> 1

C12H22O11 +12O2 --------------------> 12 CO2 + 11 H2O , DH = -5644 kJ --------->2

(1) x 2 followed by reverse

12 CO2 + 12 H2O -----------------> 2 C6H12O6 + 12 O2 , DH = +5610 kJ

C12H22O11 + 12 O2 --------------> 12 CO2 + 11 H2O , DH = -5644 kJ

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by adding above two

C12H22O11 + H2O --------------> 2 C6H12O6 , DH = -34 kJ --------------> answer

note : your statement and equation entairly different check once. this is you have given . it is wrong

Calculate the enthalpy for the reaction of hydrolysis of saccharose into two molecules of glucose:

C12H22O11(s) + H2O(l) --> 2C6H12O6(s)