Question

Ammonium nitrate dissociates in water according to the following equation:

43()→4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answer 1

Answer:

1) $q_{rxn}=1264.45J$

2) $n_{NH_4NO_3}=0.0625molNH_4NO_3$

3) $\Delta H=20.2\frac{kJ}{mol}$

Explanation:

Hello,

In this case, with the given data, we proceed as shown below:

1) q for the reaction is computed by considering the total mass of the system, that is the mass of ammonium nitrate and water:

$m=5.00g+50.0mL\times\frac{1g}{1mL} =55.0g$

Next, by using the heat capacity of the solution and the change in temperature, we obtain heat of solution:

$q_{sln}=mCp(T_2-T_1)=55.0g\times 4.18\frac{J}{g^oC}\times(16.5^oC-22.0^oC) \\\\q_{sln}=-1264.45J$

Finally, the heat of reaction:

$q_{rxn}=-q_{sln}=1264.45J$

2) For the moles of solid ammonium nitrate we use its molar mass (80.05 g/mol) and the used 5.00 g:

$n_{NH_4NO_3}=5.00gNH_4NO_3\times \frac{1molNH_4NO_3}{80.05gNH_4NO_3} \\\\n_{NH_4NO_3}=0.0625molNH_4NO_3$

3) Finally, we obtain the change in the enthalpy of reaction by using heat of reaction and the reacted moles of ammonium nitrate:

$\Delta H=\frac{q_{rxn}}{n_{NH_4NO_3}} =\frac{1264.45J}{0.0625mol}\times\frac{1kJ}{1000J}\\ \\\Delta H=20.2\frac{kJ}{mol}$

Best regards.