Question

heat of combustion of H2(g)=-241.8kj/mol. and C(s)=-393.5kj/mol. C2H5OH (l)=-1234.7kj/mol. heat of formation of C2H5OH(l) is

Answer 1

Enthalpy of formation = -1234 -(-241.8 + (-393.5)) = -1234 + 635.3 = - 598.7 kj/mol.

Answer 2

Answer : The enthalpy of formation of $C_2H_5OH(l)$ is, -277.7 KJ/mole

Solution :

The balanced chemical reaction are,

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$     $\Delta H_1=-241.8KJ/mole$  

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$    $\Delta H_2=-393.5KJ/mole$

(3) $C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l)$    $\Delta H_3=-1234.7KJ/mole$

The formation reaction of $C_2H_5OH(l)$ will be,

$2C(s)+3H_2(g)+\frac{1}{2}O_2(g)\rightarrow C_2H_5OH(l)$     $\Delta H_{formation}=?$

Now adding twice of reaction 2, thrice of reaction 1 and then subtracting reaction 3 from the addition of two reaction, we get the enthalpy of formation of $C_2H_5OH(l)$.

The expression for enthalpy of formation of $C_2H_5OH(l)$ is,

$\Delta H_{formation}=[2\times \Delta H_2]+[3\times \Delta H_1]-[1\times \Delta H_3]$

where,

n = number of moles

$\Delta H_{formation}=[2\times (-393.5)]+[3\times (-241.8)]-[1\times (-1234.7)]=-277.7KJ/mole$

Therefore, the enthalpy of formation of $C_2H_5OH(l)$ is, -277.7 KJ/mole