Question

The following two reactions are known
$F{{e}_{2}}{{O}_{3(s)}}+3C{{O}_{(g)}}\to 2Fe(s)+3C{{O}_{2(g)}};\Delta H=-26.8kJ$
$Fe{{O}_{(s)}}+C{{O}_{(g)}}\to Fe(s)+C{{O}_{2(g)}};\Delta H=-16.5kJ$
The value of $\Delta H$ for the following reaction is:
$F{{e}_{2}}{{O}_{3(s)}}+C{{O}_{(g)}}\to 2FeO(s)+C{{O}_{2(g)}}$
(a) +10.3kJ
(b) -43.3kJ
(c) -10.3kJ
(d) +6.2kJ

Answer 1

In this, we have to make the reaction whose $\Delta H$ value to be found by the addition or subtraction or multiplication of given two equations. First, we have to multiply the FeO reaction by two as the in the final equation there is two molecules of FeO and then revert that equation as FeO on the products side and then, on adding this and first equation of statement ; we get resultant equation and then, we can easily find the value of $\Delta H$ for the resultant equation.

Complete step by step solution:
First of all, what is the enthalpy of formation? From the enthalpy of formation, we simplify the total change in the enthalpy of the reaction when 1mole of the compound is formed from its constituents’ elements.
Now, considering the numerical:
We have to find the value of $\Delta H$ for the reaction;
$F{{e}_{2}}{{O}_{3(s)}}+C{{O}_{(g)}}\to 2FeO(s)+C{{O}_{2(g)}}$---------(1)
This reaction is made by combining the above two reactions by undergoing addition and subtraction and multiplication in this theme. These reactions are as;
$F{{e}_{2}}{{O}_{3(s)}}+3C{{O}_{(g)}}\to 2Fe(s)+3C{{O}_{2(g)}};\Delta H=-26.8kJ$ ---------(2)
$Fe{{O}_{(s)}}+C{{O}_{(g)}}\to Fe(s)+C{{O}_{2(g)}};\Delta H=-16.5kJ$ -----------(3)
Now, in the equation (1), we can see that there are two molecules of $FeO$, so will multiply the whole equation (3) by 2 as:
$[Fe{{O}_{(s)}}+C{{O}_{(g)}}\to Fe(s)+C{{O}_{2(g)}}]\times 2;\text{ }\Delta H=-16.5\times 2kJ$
After the equation(3), the new equation, we get is;
$2Fe{{O}_{(s)}}+2C{{O}_{(g)}}\to 2Fe(s)+2C{{O}_{2(g)}};\text{ }\Delta H=-33kJ$----------(4)
Now from the equation(1), we can see that the FeO is on the reactant side that means we will revert the equation(4) and the new equation is as;
$2Fe(s)+2C{{O}_{2}}_{(g)}\to 2FeO(s)+2C{{O}_{(g)}};\text{ }\Delta H=33kJ$ ---------(5)
Since, the reaction is reversed , so the $\Delta H$ will become positive.
Now, adding equation (2) from equation (5), we will get equation (1) as;

& \text{ }F{{e}_{2}}{{O}_{3(s)}}+3C{{O}_{(g)}}\to {2Fe(s)}+3C{{O}_{2(g)}}\text{ }\Delta H=-26.8kJ \\\ & {\text{ }{2Fe(s)}+2C{{O}_{2}}_{(g)}\to 2FeO(s)+\text{ 2C}{{\text{O}}_{(g)}}\text{ }\Delta H=33kJ\text{ }}{F{{e}_{2}}{{O}_{3(s)}}+\text{ }C{{O}_{(g)}}\to 2FeO(s)\text{ }+\text{ C}{{\text{O}}_{2(s)}}\text{ }\Delta H=(-26.8+33)kJ=+6.2kJ\text{ }} \\\ \end{aligned}$$ So, thus the value of $\Delta H$ for the following reaction is: $$F{{e}_{2}}{{O}_{3(s)}}+C{{O}_{(g)}}\to 2FeO(s)+C{{O}_{2(g)}}\text{ }\Delta \text{H=+6}\text{.2kJ}$$ **Hence, option (d) is correct.** **Note:** The enthalpy of formation of elements which are present in their molecular forms like oxygen gas, or in any solid form etc. their standard enthalpy of formation is always taken as zero as they undergo no change in their formation.