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How would calculate the standard enthalpy change for the reaction
\[2A + B = 2C + 2D\] , if \[A = - 269{\text{ }}\;B = - 411{\text{ }}C = {\text{ }}189{\text{ }}D = - 481\] ?

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: The enthalpy is the heat change of a reaction. The standard enthalpy of a reaction is the overall heat change when a reaction occurs between a set of reactants.

Complete answer:
The standard enthalpy of a reaction is defined as the heat change for the transformation of substances from one state to another or conversion of reactants into products. The standards enthalpy change of a reaction is denoted as $\Delta H^\circ .$
The standard enthalpy change of a reaction is expressed by using a mathematical equation as
\[\Delta H{^\circ _r} = \sum\limits_A {{x_A}} \Delta H{^\circ _{{f_A}}}\] where \[\Delta H{^\circ _r}\] is the standard enthalpy of reaction, \[A\] is the reference substance, \[{x_A}\] is the stoichiometric factor of \[A\] , and \[\Delta H{^\circ _{{f_A}}}\] is the standard enthalpy of formation of \[A\] .
For a general reaction, in which a reactant \[A\] is converted into product \[B\] , the standard enthalpy of the reaction is represented as
\[A \to B,{\text{ }}\Delta H^\circ = \Delta H{^\circ _{f(}}_{product)} - \Delta H{^\circ _{f(reac\tan t)}} = \Delta H{^\circ _f}_{_B} - \Delta H{^\circ _{{f_A}}}\]
Thus the enthalpy change of any reaction is determined from the above equation using the given standard enthalpy of formation of the reactants and products.
The given reaction is \[2A + B = 2C + 2D\] , where two moles of \[A\] react with one mole of \[B\] and produces two moles of \[C\] and two moles of \[D\] . Given the heat of formation of \[A,{\text{ }}B,{\text{ }}C{\text{ }}and{\text{ }}D\] are \[ - 269,{\text{ }} - 411,{\text{ }}189,{\text{ }} - 481\] respectively. The unit of heat of formation is expressed as \[KJ/mol\] or\[Kcal/mol\] .
Let us assume the given values are in \[KJ/mol\] so the standard enthalpy change for the reaction is
\[\Delta H{^\circ _{rxn}} = [2\Delta H{^\circ _{f(C)}} + 2\Delta H{^\circ _{f(D)}}] - [2\Delta H{^\circ _{f(A)}} + \Delta H{^\circ _{f(B)}}] \]
\[\Delta H{^\circ _{rxn}} = [2 \times 189 + 2 \times ( - 481)] - [2 \times ( - 269) + ( - 411)] \]
\[\Delta H{^\circ _{rxn}} = [378 - 962] - [ - 538 - 411] \]
\[\Delta H{^\circ _{rxn}} = 365KJ/mol.\]

Note:
The enthalpy change is referred to as the amount of heat which is either absorbed or evolved during the transformation of the reactants into the products at a given temperature and pressure. The positive sign of the heat of reaction indicated that heat is absorbed in the reaction.
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