Question

Standard heats of formation for reactants and products in the reaction below are provided. 2 HA(aq) + MX2(aq) → MA2(aq) + 2 HX(l) Substance ΔHf° (kJ/mol) HA(aq) -357.05 HX(l) -449.579 MA2(aq) -63.958 MX2(aq) 69.602 What is the standard enthalpy of reaction, in kJ? Report your answer to three digits after the decimal.

Answer 1

Answer : The standard enthalpy of reaction is, -318.618 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2HA(aq)+MX_2(aq)\rightleftharpoons MA_2(aq)+2HX(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(MA_2)}\times \Delta H^o_f_{(MA_2)})+(n_{(HX)}\times \Delta H^o_f_{(HX)})]-[(n_{(HA)}\times \Delta H^o_f_{(HA)})+(n_{(MX_2)}\times \Delta H^o_f_{(MX_2)})]$

We are given:

$\Delta H^o_f_{(HA(aq))}=-357.05kJ/mol\\\Delta H^o_f_{(MX_2(aq))}=69.602kJ/mol\\\Delta H^o_f_{(MA_2(aq))}=-63.958kJ/mol\\\Delta H^o_f_{(HX(l))}=-449.579kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -63.958)+(2\times -449.579)]-[(2\times -357.05)+(1\times 69.602)]=-318.618kJ$

Thus, the standard enthalpy of reaction is, -318.618 kJ