Question

The heat of combustion of sucrose at constant volume is -1348.9 kcal /mol at 25°C then the heat of reaction at constant pressure, when steam is produced is? Pls explain how to solve.

Answer 1

Please refer attached file

Answer 2

Answer : The heat of reaction at constant pressure is, -1342.409 Kcal/mole.

Explanation : Given,

Heat of combustion at constant volume = -1348.9 Kcal/mole

Temperature = $25^oC=273+25=298K$

First we have to calculate the change in moles of gaseous part.

The balanced combustion reaction will be,

$C_{12}H_{22}O_{11}(s)+12O_2(g)\rightarrow 12CO_2(g)+11H_2O(g)$

$\Delta n=n_{product}-n_{reactant}$

where,

$\Delta n$ = change in moles of gaseous part

$n_{product}$ = number of moles of product gas = 12 + 11 = 23

$n_{reactant}$ = number of moles of reactant gas = 12

$\Delta n=23-12=11mole$

Now we have to calculate the heat of reaction at constant pressure.

The relation between heat of combustion at constant volume and heat of reaction at constant pressure are :

$\Delta H=\Delta U+\Delta nRT$

where,

$\Delta H$ = heat of reaction at constant pressure

$\Delta U$ = heat of reaction at constant volume

$\Delta n$ = change in moles of gaseous part

R = gas constant = $1.98\times 10^{-3}Kcal$

T = temperature = 298 K

Now put all the given values in the above relation, we get :

$\Delta H=-1348.9Kcal/mole+11\times 1.98\times 10^{-3}Kcal\times 198K$

$\Delta H=-1342.409Kcal/mole$

Therefore, the heat of reaction at constant pressure is, -1342.409 Kcal/mole.