A drop in water temperature is observed when 0.848 g of ammonium nitrate is added. The enthalpy change for this reaction is +0.117 kJ. Calculate the Δ Hrxn, in kJ/mol, for this reaction.

The Δ Hrxn, in kJ/mol, for this reaction= 11.04 kJ/mol

Given

0.848 g ammonium nitrate

The enthalpy change : +0.117 kJ

Required

Δ Hrxn, in kJ/mol

Solution

Delta H reaction (ΔH) is the amount of heat change between the system and its environment  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

mass of NH4NO3 = 0.848 g

mol of NH4NO3(MW = 80 g/mol)

mol = 0.848 : 80

mol = 0.0106

The Δ Hrxn :

= 0.117 kJ : 0.0106 mol

= 11.04 kJ/mol

sure what is it? ^w^

a graduated cylinder.

explanation: