CBSE Class 11-science Answered

The required reaction is:

6C_(s) + 6H_2(g) + 3O_2(g)  → C₆H₁₂O_6(s)

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

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

C₆H₁₂O_{6 (s)} + 6O_2(g)  → 6CO_2(g) + 6H₂O_(l)      ΔH = -2802 kJ/mol                  …(i)

Enthaply of formation of carbondioxide is 394 kJ/mol.

C_(S) + O_2(g)  → CO_2(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,

6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_{6 (s)} + 6O_2(g)       ΔH = 2802 kJ/mol                   …(iv)

6C_(S) + 6O_2(g)  → 6CO_2(g)  ΔH = 2364 kJ/mol                                                …(v)

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

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

6C_(s) + 6H_{2 (g)} + 3O_2(g)  → C₆H₁₂O_{6 (s)}            ΔH = 2802 + 2364 + 2316 = 7482 kJ/mol

Enthalpy of formation of glucose = 7482 kJ/mol