Question

Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of Ca(OH)2 is -986.09 kJ/mol and liquid H2O is -285.8 kJ/mol.) Ca(s) + 2H2O(l) ---> Ca(OH)2 (s) + H2(g)

Answer 1

1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0

Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf [Ca(s)] were 0 because these are elements in their standard states.

HOPE THIS HELPS ;)

Answer 2

Answer : The enthalpy change for this reaction is, -414.49 KJ

Solution :

The balanced chemical reaction is,

$Ca(s)+2H_2O(l)\rightarrow Ca(OH)_2(s)+H_2(g)$

The expression for enthalpy change is,

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H_{rxn}=[(n_{H_2}\times \Delta H_{H_2})+(n_{Ca(OH)_2}\times \Delta H_{Ca(OH)_2})]-[(n_{Ca}\times \Delta H_{Ca})+(n_{H_2O}\times \Delta H_{H_2O})]$

where,

n = number of moles

As we know that standard enthalpy of an element and standard gas always be zero.

$\Delta H_{rxn}=[(n_{Ca(OH)_2}\times \Delta H_{Ca(OH)_2})]-[(n_{H_2O}\times \Delta H_{H_2O})]$

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

$\Delta H_{rxn}=[(1mole\times -986.09KJ/mole)]-[(2mole\times -285.8KJ/mole)]\\\\\Delta H_{rxn}=-414.49KJ$

Therefore, the enthalpy change for this reaction is, -414.49 KJ