Question

g The following reaction is the first step in the production of nitric acid from ammonia. 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g) Calculate the standard enthalpy change, in kJ, for this reaction

Answer 1

Answer: The enthalpy of the reaction is coming out to be -902 kJ.

Explanation:

Enthalpy change 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 $\Delta H$

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

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

For the given chemical reaction:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

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

$\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]$

We are given:

$\Delta H_f_{(NO(g))}=91.3kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-45.9kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ$

Hence, the enthalpy of the reaction is coming out to be -902 kJ.