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

Last updated date: 29th Feb 2024
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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.

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.