contestada

Ammonia (NH3(g), Delta. Hf = –46. 19 kJ/mol) reacts with hydrogen chloride (HCl(g), Delta. Hf = –92. 30 kJ/mol) to form ammonium chloride (NH4Cl(s), Delta. Hf = –314. 4 kJ/mol) according to this equation: NH3(g) HCl(g) Right arrow. NH4Cl(s) What is Delta. Hrxn for this reaction? kJ.

Respuesta :

The change in enthalpy of the given reaction is 175.91 kJ/mol, which shows that reaction is exothermic.

How we calculate ΔH for the reaction?

Change in enthalpy for the overall reaction will be calculated as:

ΔH = Enthalpy (H) of the product - Enthalpy (H) of the reactant

Given chemical reaction is:

NH₃(g) + HCl(g) → NH₄Cl

Heat of enthalpy of NH₃ = -46. 19 kJ/mol

Heat of enthalpy of HCl = -92. 30 kJ/mol

Heat of enthalpy of NH₄Cl = -314. 4 kJ/mol

Enthalpy of the reaction is calculated as:

ΔH = -314. 4 - [-46. 19 + (-92. 30)]

ΔH = -314. 4 - (-138.49)

ΔH = -175.91 kJ/mol

Hence, ΔH for the reaction is -175.91 kJ/mol.

To know more about enthalpy, visit the below link:

https://brainly.com/question/14291557

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