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# 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$kJ/mol$

Last updated date: 11th Jun 2024
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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.