Question

For which one of the following equations is $\Delta {{H}^{o}}$ (reaction) equal to $\Delta {{H}^{o}}$ (formation) for the product?
(A)${{N}_{2}}(g)+{{O}_{3}}(g)\to {{N}_{2}}{{O}_{3}}(g)$
(B) $C{{H}_{4}}\left( g \right)+2C{{l}_{2}}\left( g \right)\to C{{H}_{2}}C{{l}_{2}}\left( g \right)+2HCl\left( g \right)$
(C)$Xe\left( g \right)+2{{F}_{2}}\left( g \right)\to Xe{{F}_{4}}\left( g \right)$
(D) $2CO\left( g \right)+{{O}_{2}}\left( g \right)\to 2C{{O}_{2}}\left( g \right)$

Answer 1

Attempt this question by understanding the definition of heat of reaction and heat of formation. Heat of reaction or enthalpy of reaction is the change in the enthalpy value of a chemical reaction at constant pressure. Heat of formation or enthalpy of formation is the enthalpy change in the formation of 1 mole of a compound from its component elements. Hence we will apply these definitions to each of the given reactions to find the answer.

Complete answer:
Let us first understand the relation between heat of reaction and heat of formation:-
-”The heat of reaction (also known as enthalpy of reaction) is the change in the value of enthalpy of a chemical reaction that occurs at a constant pressure whereas the heat of formation is described as the heat generated or absorbed when 1 mole of compound is formed from its component elements in their standard state”.
-The heat of reaction for a chemical reaction can be calculated as:-
$\Delta {{H}^{{\mathrm O}}}=\sum \Delta {{n}_{g}}\Delta H_{f}^{{\mathrm O}}(products)-\sum \Delta {{n}_{g}}\Delta H_{f}^{{\mathrm O}}(reactants)$
where,
$\begin{aligned} & \Delta {{n}_{g}}=\text{stoichiometric coefficients in the balanced equation} \\\ & \Delta H_{f}^{{\mathrm O}}=\text{heat of formation} \\\ \end{aligned}$
-The elements which are already in their standard states have zero enthalpy of formation. So now we can apply these above definitions in the given reactions as follows:-
(A)${{N}_{2}}(g)+{{O}_{3}}(g)\to {{N}_{2}}{{O}_{3}}(g)$:-
Heat of reaction $\Delta {{H}^{\text{O}}}=\Delta H_{f}^{\text{O}}({{N}_{2}}{{O}_{3}}(g))-\\{\Delta H_{f}^{\text{O}}({{N}_{2}}(g))+\Delta H_{f}^{\text{O}}({{O}_{3}}(g))\\}$
Since we know that the standard state of nitrogen is ${{N}_{2}}(g)$, therefore its enthalpy of formation is zero whereas ${{O}_{3}}(g)$ is not the standard state of oxygen, so its enthalpy of formation is not zero.
$\Rightarrow \Delta {{H}^{\text{O}}}=\Delta H_{f}^{\text{O}}({{N}_{2}}{{O}_{3}}(g))-\Delta H_{f}^{\text{O}}({{O}_{3}}(g))$
Hence for this reaction $\Delta {{H}^{o}}$ (reaction) is not equal to $\Delta {{H}^{o}}$ (formation) for the product.
(B) $C{{H}_{4}}\left( g \right)+2C{{l}_{2}}\left( g \right)\to C{{H}_{2}}C{{l}_{2}}\left( g \right)+2HCl\left( g \right)$
Heat of reaction $\Delta {{H}^{\text{O}}}=\\{\Delta H_{f}^{\text{O}}(C{{H}_{2}}C{{l}_{2}}(g))+2\Delta H_{f}^{\text{O}}(HCl(g))\\}-\\{\Delta H_{f}^{\text{O}}(C{{H}_{4}}(g))+2\Delta H_{f}^{\text{O}}(C{{l}_{2}}(g))\\}$
Since we know that the standard state of chlorine is ${{Cl}_{2}}(g)$, therefore its enthalpy of formation is zero whereas $C{{H}_{4}}(g)$is a compound which can be formed by other elements, so its enthalpy of formation is not zero.
$\Rightarrow \Delta {{H}^{\text{O}}}=\\{\Delta H_{f}^{\text{O}}(C{{H}_{2}}C{{l}_{2}}(g))+2\Delta H_{f}^{\text{O}}(HCl(g))\\}-\\{\Delta H_{f}^{\text{O}}(C{{H}_{4}}(g))\\}$
Hence for this reaction $\Delta {{H}^{o}}$ (reaction) is not equal to $\Delta {{H}^{o}}$ (formation) for the product.
(C)$Xe\left( g \right)+2{{F}_{2}}\left( g \right)\to Xe{{F}_{4}}\left( g \right)$
Heat of reaction $\Delta {{H}^{\text{O}}}=\Delta H_{f}^{\text{O}}(Xe{{F}_{4}}\left( g \right))-\\{\Delta H_{f}^{\text{O}}(Xe\left( g \right))+2\Delta H_{f}^{\text{O}}({{F}_{2}}\left( g \right))\\}$
Since xenon and fluorine are present in their standard state, therefore their enthalpy of formation is zero.
$\Rightarrow \Delta {{H}^{\text{O}}}=\Delta H_{f}^{\text{O}}(Xe{{F}_{4}}\left( g \right))$
Hence for this reaction $\Delta {{H}^{o}}$ (reaction) is equal to $\Delta {{H}^{o}}$ (formation) for the product.
(D) $2CO\left( g \right)+{{O}_{2}}\left( g \right)\to 2C{{O}_{2}}\left( g \right)$
Heat of reaction $\Delta {{H}^{\text{O}}}=2\Delta H_{f}^{\text{O}}(C{{O}_{2}}\left( g \right))-\\{2\Delta H_{f}^{\text{O}}(CO\left( g \right))+\Delta H_{f}^{\text{O}}({{O}_{2}}(g))\\}$
Since oxygen is present in its standard state, therefore its enthalpy of formation is zero whereas CO (g) is a compound which can be formed by other elements, so its enthalpy of formation is not zero.
$\Rightarrow \Delta {{H}^{\text{O}}}=2\Delta H_{f}^{\text{O}}(C{{O}_{2}}\left( g \right))-2\Delta H_{f}^{\text{O}}(CO\left( g \right))$
Hence for this reaction $\Delta {{H}^{o}}$ (reaction) is not equal to $\Delta {{H}^{o}}$ (formation) for the product.

-From the above data we conclude that the correct answer is: (C)$Xe\left( g \right)+2{{F}_{2}}\left( g \right)\to Xe{{F}_{4}}\left( g \right)$

Note:
-While solving such questions always check the standard state of the elements along with their phases because standard phase is also considered.
-We can also say that $\Delta {{H}^{o}}$ (reaction) is equal to $\Delta {{H}^{o}}$ (formation) for the product when 1 mole of compound is formed from its constituent element (which are required to be in their standard states).