Question

A student dissolves 12.6g of amonium nitrate(NH4NO3) in 250.g of water in a well-insulated open cup. She then observed the temperature of the water fall from 23.0°C to 18°C over the course of 6.1 minutes.

NH4NO3 â NH4+ (aq) + NO3^-(aq)

a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or calculate the amount of heat that was released or absorbed by the reaction in this case.
c. Calculate the reaction enthalpy ÎHrxn per mole of NH4NO3.

Answer 1

a. Endothermic.

b.
`Q_(rxn)=5493.6J`

c.
`\Delta H_(rxn)=35.0kJ/mol`

Step-by-step explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, it turns out possible for us to proceed as follows:

a. Due to the fact that the temperature of water goes from 23.0 °C to 18.0 °C, we infer this reaction is endothermic as the ammonium nitrate absorbed heat from the water.

b. Here, we consider the following heat equation:

`Q_(rxn)=-Q_(water)`

Whereas we solve for the heat of reaction by means of the mass of the solution (both water and ammonium nitrate), the specific heat of the solution (we assume it is equal to that of water) and the temperature change:

`Q_(rxn)=-m_(solution)C_(solution)(T_f-T_i)\\\\Q_(rxn)=-(12.6g+250.g)(4.184(J)/(g\°C) )(18.0\°C-23.0\°C)\\\\Q_(rxn)=5493.6J`

c. Here, we divide the previously calculated heat by the moles of ammonium nitrate (molar mass = 80.043 g/mol) to obtain the enthalpy of reaction per mole of this compound:

`n_(NH_4NO_3)=12.6g*(1mol)/(80.043 g)=0.157mol\\\\\Delta H_(rxn)=(5493.6J)/(0.157mol) =34898.7J/mol\\\\\Delta H_(rxn)=35.0kJ/mol`

Regards!