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

Respuesta :

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 [tex]\Delta H[/tex]

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

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

For the given chemical reaction:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

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

[tex]\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)})][/tex]

We are given:

[tex]\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[/tex]

Putting values in above equation, we get:

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

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

The enthalpy of reaction for the first step of the production of nitric acid from ammonia is -902 kJ/mol.

Standard enthalpy change

The standard enthalpy change of a reaction cal be calculated using the enthapies of formation of the reactants and the products in the reaction. Recall that the enthalpy of formation of a pure compound is 0 Kj/mol.

Enthalpy of formation of ammonia = -45.9 kJ/mol

Enthalpy of formation of nitogen monoxide = 91.3 kJ/mol

Enthalpy of formation of gaseous water = -241.8 kJ/mol

Enthalpy of formation of oxygen = 0 kJ/mol

ΔHrxn = [4(91.3 kJ/mol) + 6( -241.8 kJ/mol)] - [4(-45.9 kJ/mol) + 5( 0 kJ/mol)]

ΔHrxn = [365.2 - 1450.8] + 183.6 = -902 kJ/mol

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