Question

Change in enthalpy for reaction:
\[2{H_2}{O_2}\left( l \right) \to 2{H_2}O\left( l \right) + {O_2}\left( g \right)\]. If heat of formation of \[{H_2}{O_2}\left( l \right)\] and\[{H_2}O\left( l \right)\] are -188 and -286 \[kJ/mol\]respectively:
(A) -196\[kJ/mol\]
(B) +196\[kJ/mol\]
(C) +948\[kJ/mol\]
(D) -948\[kJ/mol\]

Answer 1

Hint - In this question we will come across the concept of thermodynamics. In this question we will be crossing path with many important concepts like enthalpy, reaction enthalpy\[{\Delta _r}H\]and about standard enthalpy of formation\[{\Delta _f}{H^\Theta }\]. And this information will help us in approaching our answer. Below here we have explained each topic very properly.

Complete step by step solution:
> Enthalpy: We know that energy change occurring during the reaction at a constant pressure and constant volume is given by internal energy change that is, heat absorbed at constant volume is equal to change in internal energy that is\[\Delta U = {q_v}\]. As atmospheric pressure is constant, therefore, such reactions may involve change in volume. It is a sum of internal energy and pressure-volume energy of the system at a particular temperature and pressure.
\[\Delta H\]Enthalpy change is the measure of heat change taking place during a process at constant temperature and pressure.
> Reaction enthalpy\[{\Delta _r}H\]: the enthalpy change accompanying a chemical reaction when the number of moles of reactants react to give the products as by the balanced chemical equation.
Standard enthalpy of formation\[{\Delta _f}{H^\Theta }\]: The enthalpy change accompanying the formation of one mole of the compounds from its elements at standard conditions and all the substance being their standard states.
As a convection\[{\Delta _f}{H^\Theta }\] of every element is assumed to be zero.
\[2{H_2}{O_2}\left( l \right) \to 2{H_2}O\left( l \right) + {O_2}\left( g \right)\]
\[{\Delta _f}{H^\Theta }\][\[{H_2}{O_2}\left( l \right)\]] = -188\[kJ/mol\]
For 2 moles \[{H_2}{O_2}\left( l \right)\] = -2 \[ \times \]188\[kJ/mol\]
\[{\Delta _f}{H^\Theta }\][\[{H_2}{O_2}\left( l \right)\]] = -286\[kJ/mol\]
For 2 moles of \[{H_2}O\left( l \right)\] = -2 \[ \times \]286\[kJ/mol\]
\[{\Delta _f}{H^\Theta }\]= (2\[ \times \]-286) – (2\[ \times \] -188)
= -196\[kJ/mol\]

Hence, the correct answer is option (A).

Note - In this question we have learned about enthalpy, about reaction enthalpy\[{\Delta _r}H\]and about standard enthalpy of formation\[{\Delta _f}{H^\Theta }\]. Change in enthalpy is measured by calorimeter and the process is called calorimeter. This information we have learned in this question will help us in future.