Question

M is a metal that forms an oxide ${M_2}O$,
$\dfrac{1}{2}{M_2}O \to M + \dfrac{1}{4}{O_2},{\text{ }}\Delta H = 120 KCal$
When a sample of metal $M$ reacts with one mole of oxygen what will be the $\Delta H$ in that case
A. $240Kcal$
B. $- 240Kcal$
C. $480Kcal$
D. $- 480KCal$

Answer 1

The heat of a reaction or the enthalpy change of the reaction is measured as h. It is equal to the difference of enthalpy of products and enthalpy of reactants.

Complete step by step answer:
The metal $M$ upon oxidation or reaction with oxygen produces a metal oxide ${M_2}O$, where ${M_2}O$ is the chemical formula of the metal oxide.
The corresponding equation given is the decomposition of ${M_2}O$ to generate metal $M$ and oxygen which is
$\dfrac{1}{2}{M_2}O \to M + \dfrac{1}{4}{O_2},{\text{ }}\Delta H = 120 KCal$
The reaction is an endothermic reaction as the value of change in enthalpy is a positive number. If we reverse the reaction, i.e. the formation of M2O from the reaction of M and oxygen, the value of enthalpy change becomes negative. The reaction is
$M + \dfrac{1}{4}{O_2} \to \dfrac{1}{2}{M_2}O,{\text{ }}\Delta H = - 120 KCal.$
From stoichiometry of the reaction, it is clear that one equivalent or mole of metal $M$ reacts with $\dfrac{1}{4}$ equivalents or moles of oxygen to produce $\dfrac{1}{2}$ moles of ${M_2}O$.
Now, if according to the question one mole of oxygen is reacted with metal $M$, the enthalpy change for the reaction will change. The corresponding reaction is multiplied by $4$,
$4\left( {\dfrac{1}{4}} \right){O_2} + 4M \to 4\left( {\dfrac{1}{2}} \right){M_2}O$
${O_2} + 4M \to 2{M_2}O$
Thus from the modified equation, one mole of oxygen reacts with $4$ moles of metal $M$ to generate $2$ moles of ${M_2}O$.
Therefore multiplying the change in enthalpy by $4$ gives the new enthalpy change.
$\Delta H = 4 \times ( - 120KCal) = - 480KCal$.
So, the correct answer is “Option D”.

Note:
The enthalpy change of a reaction is the amount of heat absorbed or released of a reaction provided that the pressure during the transformation is constant. It only applies to reactions where pressure is fixed and the standard enthalpy change is at $1bar\;\left( {100kPa} \right)$ pressure.