Question

Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.

Answer 1

Answer: -134 kJ

Explanation:

The balanced chemical reaction is,

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)$

The expression for enthalpy change is,

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

$\Delta H=[(n_{HNO_3}\times \Delta H_{HNO_3})+(n_{NO}\times \Delta H_{NO})]-[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO_2}\times \Delta H_{NO_2})]$

where,

n = number of moles

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

$\Delta H=[(2\times -207)+(1\times 90)]-[(1\times -286)+(3\times 32)]$

$\Delta H=-134kJ$

Therefore, the enthalpy change for this reaction is, -134 kJ