In a coffee-cup calorimeter, 1.91 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00oC. After the solution of the salt, the final temperature was 20.94oC. Assuming the solution has a heat capacity of 4.18 J/(goC) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.
q soln = mass H2O x specific heat H2O x (Tfinal-Tinitial). This give you q for 1.91g NH4NO3.
(q/1.91g) x molar mass NH4NO3 gives you J/mol, then convert to kJ/mol.
Plug in the numbers to obtain q
52.9kj/mole
To calculate the enthalpy change for the dissolution of NH4NO3, we need to use the heat transfer formula:
q = m * c * ΔT
where:
q is the heat transfer (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g°C)), and
ΔT is the change in temperature (in °C).
First, let's calculate the heat transferred during the dissolution:
Mass of NH4NO3 (mNH4NO3) = 1.91 g
Specific heat capacity of the solution (c) = 4.18 J/(g°C)
Change in temperature (ΔT) = Final temperature - Initial temperature = 20.94°C - 25.00°C = -4.06°C
Now, substitute the values into the formula:
q = mNH4NO3 * c * ΔT
= 1.91 g * 4.18 J/(g°C) * -4.06°C
= -32.1644 J
Next, to convert the heat transferred from joules to kilojoules, we divide the value by 1000:
q = -32.1644 J / 1000
= -0.0321644 kJ
Since the amount of NH4NO3 used is given in grams, we need to convert it to moles by using the molar mass of NH4NO3.
Molar mass of NH4NO3 = (1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol)
= 80.04 g/mol
Moles of NH4NO3 = Mass of NH4NO3 / Molar mass of NH4NO3
= 1.91 g / 80.04 g/mol
= 0.023862 moles
Finally, to calculate the enthalpy change (ΔH) in kJ/mol, divide the heat transferred (q) in kJ by the number of moles of NH4NO3:
ΔH = q / Moles of NH4NO3
= -0.0321644 kJ / 0.023862 moles
≈ -1.347 kJ/mol
Therefore, the enthalpy change for the dissolution of NH4NO3 is approximately -1.347 kJ/mol.