Question

Using the standard heats of formation that follow, calculate the standard enthalpy change for the following reaction:Fe(s) + 2HCl(aq)---->FeCl2(s) + H2(g) deltaH rxn=________kJSpecies H°f (kJ/mol)HCl(aq) -167.2FeCl2(s) -341.8

Answer 1

Answer: -7.4 kJ

Explanation:

The balanced chemical reaction is:

$Fe(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{FeCl_2}\times \Delta H_{FeCl_2})+(n_{H_2}\times \Delta H_{H_2})]-[(n_{Fe}\times \Delta H_{Fe})+(n_{HCl}\times \Delta H_{HCl})]$

where,

n = number of moles

$\Delta H_{Fe}=0$ and $\Delta H_{H_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$\Delta H=[(1\times -341.8)+(1\times 0)]-[(1\times 0)+(2\times -167.2)]$

$\Delta H=-7.4kJ$

Therefore, the enthalpy change for this reaction is -7.4 kJ