When a 3.90-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 24.5°C to 19.3°C. Calculate DH (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.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = heat loss by the water in J. Divide by 1000 to convert to kJ. That gives you total kJ. Then kJ/3.90 gives kJ/gram NH4NO3 and
(kJ/3.90)*molar mass NH4NO3 gives kJ/mol.
1305.41
26.78
To calculate ΔH (enthalpy change) for the solution process, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change (in kJ/mol NH4NO3), q is the heat absorbed or released (in kJ), and n is the number of moles of NH4NO3.
First, let's calculate q:
q = m * C * ΔT
where q is the heat absorbed or released (in kJ), m is the mass of the solution (in g), C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (in °C).
Given:
- Mass of the solution (m) = mass of NH4NO3 + mass of water = 3.90 g + 60.0 g = 63.9 g
- Specific heat capacity of water (C) = 4.18 J/g°C
- Change in temperature (ΔT) = final temperature - initial temperature = 19.3°C - 24.5°C = -5.2°C
Converting the values to joules (1 kJ = 1000 J):
m = 63.9 g = 63.9 g * 1 kJ/1000 J = 0.0639 kg
ΔT = -5.2°C = -5.2°C * 1 K/°C = -5.2 K
q = 0.0639 kg * 4.18 J/g°C * -5.2 K = -1.377264 J
Converting q to kJ:
q = -1.377264 J / 1000 J/kJ = -0.001377264 kJ
Now, let's calculate the number of moles (n) of NH4NO3:
n = mass / molar mass
Given:
- Mass of NH4NO3 = 3.90 g
- Molar mass of NH4NO3 = 80.04 g/mol
n = 3.90 g / 80.04 g/mol = 0.0487 mol
Finally, let's substitute the values into the ΔH equation:
ΔH = q / n = -0.001377264 kJ / 0.0487 mol = -0.0282 kJ/mol NH4NO3
Therefore, ΔH for the solution process is -0.0282 kJ/mol NH4NO3.
To calculate the enthalpy change (ΔH) for the solution process of ammonium nitrate (NH4NO3), we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat transferred in the reaction, and n is the number of moles of the substance.
To determine the heat transferred (q), we can use the formula:
q = m * c * ΔT
where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's calculate the mass of the water in the solution:
Mass of water = 60.0 g
Next, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 19.3°C - 24.5°C = -5.2°C
Since the temperature dropped, the ΔT value is negative. However, we will work with the absolute value of the ΔT for calculations.
Let's convert the mass of the water and the change in temperature to their respective SI units:
Mass of water = 60.0 g = 0.0600 kg
ΔT = -5.2°C = -5.2 K (since K and °C are the same size units)
Now, let's calculate the heat transferred (q):
q = m * c * ΔT
q = 0.0600 kg * 4.18 J/g°C * -5.2 K
q = -1.2456 J
The negative sign indicates that heat is being lost during the process.
Finally, let's calculate the number of moles of ammonium nitrate (n):
n = mass / molar mass
n = 3.90 g / (14.01 g/mol + 1.01 g/mol + 3 * 16.00 g/mol)
n = 3.90 g / 80.043 g/mol
n = 0.04872 mol
Now, we can calculate the enthalpy change (ΔH):
ΔH = q / n
ΔH = -1.2456 J / 0.04872 mol
ΔH = -25.51 J/mol
To convert the answer to kilojoules per mole (kJ/mol), we divide by 1000:
ΔH = -25.51 J/mol / 1000
ΔH = -0.0255 kJ/mol
Therefore, the enthalpy change (ΔH) for the solution process of NH4NO3 is approximately -0.0255 kJ/mol.