Question

At 17 degree Celsius, the enthalpy of combustion of graphite to CO2 is -23110J and that of CO to CO2 is -16261J. Determine the enthalpy of formation of CO at 17 degree Celsius.

Answer 1

A given mass of gas occupies 200 ml at 27 degree Celsius volume at will accupi at 27 degree Celsius ?

Answer 2

Answer : The enthalpy of formation of CO is, 6849 J

Solution :

Using Hess's law :

The balanced chemical reaction are,

(1) $C(graphite)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$    $\Delta H_1=-23110J$

(2) $CO+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$    $\Delta H_2=-16261J$

The formation reaction of CO will be,

$C(graphite)+\frac{1}{2}O_2(g)\rightarrow CO(g)$    $\Delta H_{formation}=?$

The expression for enthalpy of formation of CO is,

$\Delta H_{formation}=[n\times \Delta H_1]+[n\times \Delta H_2]$

where,

n = number of moles

Now adding reaction 1 and reverse reaction of reaction 2, we get the enthalpy of formation of CO.

$\Delta H_{formation}=[1\times (-23110J)]+[1\times (16261J)]=6849J$

Therefore, the enthalpy of formation of CO is, 6849 J