chemistry!!?

Question

When a 5.00 g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from 22.0°C to 17.0°C. Calculate H (in kJ/mol NH4NO3) for the solution process shown below. Assume that the specific heat of the solution is the same as that of pure water.

NH4NO3(s) ------> NH4+(aq) + NO3-(aq)

∆H =_______kJ/mol

note: this is an endothermic reaction

Answer 1

The guiding formula to use to solve this problem is:
Q = mcΔT
Where Q is the heat energy released/absorbed, c is the specific heat, and ΔT is the change in temperature.

We are told to assume the specific heat of the solution is the same as that of water.
c = 4.184 Joules per gram per °C.

The mass of the solution will be the combination of the mass of the water and the mass of the Ammonium Nitrate dissolved in the water.
m_total = m_water + m_NH4NO3
m_total = (60.0 g) + (5.00 g)
m_total = 65.0 grams

The change in temperature can be found by taking the difference of the initial and final temperatures.
ΔT = (22.0 °C) – (17.0 °C)
ΔT = 5.0 °C

We now know all the information we need to find Q.
Q = mcΔT
Q = (65.0 g) * (4.184 J / g °C) * (5.0 °C)
Q = 1359.8 Joules

But this amount of only for 5.00 grams of Ammonium Nitrate, we are asked to find the ΔH value per mole.
How many moles of NH4NO4 are there in 5.00 grams? To find this out, simply divide the mass of the sample (5.00 g) by the molar mass of Ammonium Nitrate (80.06 grams per mole).
Moles of NH4NO3 = (5.00 grams) / (80.06 grams per mole)
Moles of NH4NO3 = .0625 moles

So now we divide our energy (Q) we just found by the number of moles of Ammonium Nitrate to get the energy per mole.
Q / mole = (1359.8 Joules) / (.0625 moles)
Q / mole = 21756.8 Joules per mole
Q / mole = 22000 Joules per mole
Q / mole = 22 kJ per mole

And since the temperature is dropping, energy is being absorbed during this process and it is endothermic, so the sign of ΔH must be positive.
ΔH = +22 kJ per mole of Ammonium Nitrate