Ammonia (NH3(g), es001-1.jpgHf = –46.19 kJ/mol) reacts with hydrogen chloride (HCl(g), es001-2.jpgHf = –92.30 kJ/mol) to form ammonium chloride (NH4Cl(s), es001-3.jpgHf = –314.4 kJ/mol) according to this equation: NH3(g) + HCl(g) es001-4.jpg NH4Cl(s) What is es001-5.jpgHrxn for this reaction? kJ

Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.

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nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-20T03:20:56+02:00

The value of heat of the reaction for the given chemical reaction is equal to -175.91 kJ.

How do we calculate the change in enthalpy of the reaction?

Change in enthalpy of the reaction is calculated by substracting the total sum of enthalpies of reacatnts from the the total sum of the enthalpies of products.

Given chemical reaction is:

NH₃(g) + HCl(g) → NH₄Cl(s)

According to the equation total enthalpy of the reaction calculated as:

ΔHrxn = ΔfH(NH₄Cl) - [(ΔfH(NH₃) + ΔfH(HCl)]

On putting values from the question to the equation, we get

ΔHrxn = -314.4 kJ/mol - [-46.19 kJ/mol + (-92.30 kJ/mol)]

ΔHrxn = -314.4 kJ + 138.49 kJ.

ΔHrxn = -175.91 kJ

Hence the heat of the reaction is -175.91 kJ.

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https://brainly.com/question/18721983