(a) When a 4.75-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (see figure), the temperature drops from 23.0�‹C to 16.6�‹C. Calculate ƒ¢H (in kJ/mol NH4NO3) for the solution process
NH4NO3(s) �¨ NH4+(aq) + NO3−(aq).
Assume that the specific heat of the solution is the same as that of pure water.
To calculate the enthalpy change (ΔH) for the solution process, we need to apply the equation:
ΔH = q / n
Where:
ΔH = enthalpy change (in kJ/mol)
q = heat exchanged in the reaction (in J)
n = number of moles of the substance involved in the reaction
To find the value of q, we can use the equation:
q = mcΔT
Where:
m = mass of the water (in g)
c = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (in °C)
First, let's calculate the heat exchanged (q):
q = mcΔT
= 60.0 g × 4.18 J/g·°C × (16.6�‹C - 23.0�‹C)
= -1722.24 J
Next, we need to determine the number of moles of NH4NO3 involved in the reaction. To do this, we'll use the molar mass of NH4NO3.
Molar mass of NH4NO3:
N = 14.01 g/mol
H = 1.01 g/mol (x 4 because there are 4 Hydrogen atoms)
O = 16.00 g/mol (x 3 because there are 3 Oxygen atoms)
Total = 80.04 g/mol
Now let's calculate the number of moles (n) of NH4NO3:
n = mass / molar mass
= 4.75 g / 80.04 g/mol
= 0.0594 mol
Finally, we can find the enthalpy change (ΔH) using the equation:
ΔH = q / n
= -1722.24 J / 0.0594 mol
≈ -28,983 J/mol
To convert J to kJ, divide by 1000:
ΔH = -28,983 J/mol / 1000
≈ -28.98 kJ/mol
Therefore, the enthalpy change for the solution process of NH4NO3 is approximately -28.98 kJ/mol.
To calculate the enthalpy change (ΔH) for the solution process of ammonium nitrate (NH4NO3) in water, we need to use the formula:
ΔH = q / n
Where:
- q is the heat absorbed or released by the system (in this case, the solution) in joules (J)
- n is the amount of substance in moles (mol) involved in the reaction
First, we need to calculate the heat absorbed or released by the system using the equation:
q = mcΔT
Where:
- q is the heat absorbed or released by the system in joules (J)
- m is the mass of the solution (in this case, the mass of water since the specific heat is the same) in grams (g)
- c is the specific heat capacity of water, which is 4.184 J/g·°C
- ΔT is the change in temperature of the solution in °C (final temperature - initial temperature)
Given data:
- Mass of ammonium nitrate (NH4NO3) = 4.75 g
- Mass of water (m) = 60.0 g
- Initial temperature of the solution = 23.0°C
- Final temperature of the solution = 16.6°C
Let's calculate the heat absorbed or released by the system first:
q = mcΔT
q = (60.0 g)(4.184 J/g·°C)(16.6 - 23.0)°C
Now, let's calculate the number of moles of ammonium nitrate (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
Number of moles of NH4NO3 = Mass of NH4NO3 / Molar mass of NH4NO3
Number of moles of NH4NO3 = 4.75 g / 80.04 g/mol
Now, we can calculate the enthalpy change (ΔH) for the solution process:
ΔH = q / n
Plug in the values:
ΔH = q / (4.75 g / 80.04 g/mol)