Calculate the enthaphy change of formation of gluocouse by applying Hess's law, given enthaphy of combustion of glucouse is equal to -2802kJ/mol and enthaply of formation od carbondioxide = 394kJ/mol and enthaply of formation of water is 386kJ/mol.

Asked by aishwarya jay | 11th Sep, 2014, 01:36: AM

Expert Answer:

The required reaction is:

6C(s) + 6H2(g) + 3O2(g)  → C6H12O6(s)

We have to find out standard enthalpy of formation for glucose.

Enthalpy of combustion of glucose is -2802 kJ/mol.

C6H12O6 (s) + 6O2(g)  → 6CO2(g) + 6H2O(l)      ΔH = -2802 kJ/mol                  …(i)

 

Enthaply of formation of carbondioxide is 394 kJ/mol.

C(S) + O2(g)  → CO2(g)         ΔH = 394 kJ/mol                                                …(ii)

 

Enthaply of formation of water is 386 kJ/mol.

 H2 + ½ O2 → H2O              ΔH = 386 kJ/mol                                                …(iii)

 

Now, revert eq. (i), multiply eq. (ii) and (iii) by 6,

6CO2(g) + 6H2O(l) → C6H12O6 (s) + 6O2(g)       ΔH = 2802 kJ/mol                   …(iv)

 

6C(S) + 6O2(g)  → 6CO2(g)  ΔH = 2364 kJ/mol                                                …(v)

 

6H2 + 3O2 → 6H2O            ΔH = 2316 kJ/mol                                                …(vi)

 

Add eq.(iv), (v), (vi) we will get,

 

6C(s) + 6H2 (g) + 3O2(g)  → C6H12O6 (s)            ΔH = 2802 + 2364 + 2316 = 7482 kJ/mol

Enthalpy of formation of glucose = 7482 kJ/mol

Answered by Arvind Diwale | 15th Sep, 2014, 11:26: AM