Discussion Forum

The standard enthalpies of formation of CO₂(g), H₂O(l)  and glucose (s)   at 25° C are -400 kJ/mol, -300kJ/mol and -1300kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° C  is 

Answer:

Explanation:

 Plan   $\triangle _{c}H^{0}$   (standard heat of combustion ) is the standard enthalpy change when one mole of the substance is completely oxidised.

  Also standard heat of formation ( $\triangle _{f}H^{0}$ ) can be taken as the standard  of their substance.

   $H^{0}_{H_{2}O}=\triangle_{f}H^{0}(H_{2}O)=-300kJmol^{-1}$

 $H^{0}_{glucose }=\triangle_{f}H^{0}(glucose)=-1300kJmol^{-1}$

      $H^{0}_{O_{2} }=\triangle_{f}H^{0}(O_{2})=0.00$

  $C_{6}H_{12}O_{6}(s)+6O_{2}(g)\rightarrow 6CO_{2}(g)+6H_{2}O(l)$

  $\triangle_{c}H^{0} (glucose)$

                 = $6[\triangle_{f}H^{0}(CO_{2})+\triangle_{f}H^{0} (H_{2}O)]$

       -  $[\triangle_{f}H^{0}(C_{6}H_{12}O_{6})+6\triangle_{f}H^{0} O_{2})]$

    =6[-400-300]-[-1300+6 X0]

   =-2900 kJ mol^-1

 Molar mass of C₆H₁₂O₆= 180 g mol^-1

 Thus, standard heat of combustion of glucose per gram

  =   $\frac{-2900}{180}=-16.11kJ g^{-1}$