
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
126.3k+ views
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.
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.
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