Question

The standard enthalpy of formation of ${H_2}O$(l) and $F{e_2}{O_3}$(s) are respectively -286 $kJmo{l^{ - 1}}$ and -824 $kJmo{l^{ - 1}}$. What is the standard enthalpy change for the following reaction ?
$F{e_2}{O_3}(s) + 3{H_2}(g) \to 3{H_2}O(l) + 2Fe(s)$
a.) -538 $kJmo{l^{ - 1}}$
b.) +538 $kJmo{l^{ - 1}}$
c.) -102 $kJmo{l^{ - 1}}$
d.) +34 $kJmo{l^{ - 1}}$
e.) -34 $kJmo{l^{ - 1}}$

Answer 1

Hint: We have the formula for the standard enthalpy change as -
Standard enthalpy change = (Sum of standard enthalpy of formation of products - Sum of standard enthalpy of formation of reactants)
The standard enthalpy of a reaction is thus the change in energy which occurs during change of reactants to products.

Complete answer:
Let us first understand what is standard enthalpy of formation.
It has been even named as standard heat of formation. As the word defines itself, it is the change in enthalpy or energy when one mole of a substance is formed from its constituent elements. It is usually measured in kilojoule per mole or joule per mole.
Now, coming to the question, we have been given
the standard enthalpy of formation of ${H_2}O$(l) = -286 $kJmo{l^{ - 1}}$ and
the standard enthalpy of formation of $F{e_2}{O_3}$(s) = -824 $kJmo{l^{ - 1}}$
The reaction is :-

$F{e_2}{O_3}(s) + 3{H_2}(g) \to 3{H_2}O(l) + 2Fe(s)$

Further, we know that the standard enthalpy of formation of ${H_2}$ and Fe is zero.
We have to find the standard enthalpy change for the reaction.
Standard enthalpy change = (Sum of standard enthalpy of formation of products - Sum of standard enthalpy of formation of reactants)
Standard enthalpy change = (-286 × 3) - (-824 × 1)
Standard enthalpy change = (-858) + 824
Standard enthalpy change = -34 $kJmo{l^{ - 1}}$
Thus, the value for the standard enthalpy change comes out to be -34 $kJmo{l^{ - 1}}$.

So, the option e.) is the correct answer.

Note: It must be noted that the change can be positive or negative. If the energy is consumed in the reaction then the value comes out to be positive and the reaction is endothermic in nature and in case the heat is released then the value comes out to be negative and is an exothermic reaction.
The units while subtraction should be the same. Further, the reactant and product values of each type of the molecule is to be multiplied by coefficients present in balanced reaction.