A student dissolved 10.5 g of NH4Cl in 40 mL of water. The temperature of the water changed from 22.5oC to 5.1oC. Calculate the enthalpy of hydration of ammonium chloride in J/mol.
Well, it looks like this student is having a chilling experience with ammonium chloride. Let's dive into the question!
To calculate the enthalpy of hydration, we need to use the formula:
ΔH = q / n
where ΔH represents the enthalpy change, q represents the amount of heat absorbed or released, and n represents the number of moles of the substance.
First, we need to calculate the amount of heat absorbed or released (q). We can do this using the equation:
q = m × C × ΔT
where q is the heat absorbed or released, m is the mass of the substance, C is the specific heat capacity of water, and ΔT is the change in temperature.
Substituting the given values:
m = 10.5 g (the mass of NH4Cl)
C = 4.18 J/g°C (the specific heat capacity of water)
ΔT = (22.5 - 5.1) °C (the change in temperature)
q = 10.5 g × 4.18 J/g°C × (22.5 - 5.1) °C
Now, let's calculate the number of moles (n) of NH4Cl using its molar mass.
Molar mass of NH4Cl = 14.01 g/mol for nitrogen + 1.01 g/mol for hydrogen + 4 × 35.45 g/mol for chlorine
Now we can calculate the number of moles:
n = 10.5 g / (14.01 g/mol + 4 × 1.01 g/mol + 4 × 35.45 g/mol)
Finally, we can substitute the values of q and n into the formula for enthalpy change:
ΔH = q / n
And that's how you calculate the enthalpy of hydration! Keep in mind that this is just a basic explanation, but I hope it helps.
To calculate the enthalpy of hydration of ammonium chloride (NH4Cl) in J/mol, we need to use the equation:
q = m * c * ΔT
Where:
- q is the heat absorbed or released during the process (in J)
- m is the mass of the water (in grams)
- c is the specific heat capacity of water (4.18 J/g°C)
- ΔT is the change in temperature (in °C)
Step 1: Calculate the heat absorbed by the water:
q = m * c * ΔT
First, we need to convert the volume of water (40 mL) to grams:
Density of water = 1 g/mL
Volume of water = 40 mL
Mass of water = Volume of water * Density of water
Mass of water = 40 mL * 1 g/mL
Mass of water = 40 g
Now, we can calculate the heat absorbed by the water:
q = m * c * ΔT
q = 40 g * 4.18 J/g°C * (5.1 - 22.5) °C
q = 40 g * 4.18 J/g°C * -17.4 °C
q = -2870.4 J
Step 2: Calculate the moles of NH4Cl:
Molar mass of NH4Cl = 14.01 g/mol (molar mass of NH4) + 1.01 g/mol (molar mass of H) + 35.45 g/mol (molar mass of Cl)
Molar mass of NH4Cl = 53.46 g/mol
To calculate the moles of NH4Cl, we can use the equation:
moles = mass / molar mass
moles of NH4Cl = 10.5 g / 53.46 g/mol
moles of NH4Cl = 0.1966 mol
Step 3: Calculate the enthalpy of hydration:
Enthalpy of hydration = q / moles of NH4Cl
Enthalpy of hydration = -2870.4 J / 0.1966 mol
Enthalpy of hydration ≈ -14609 J/mol
Therefore, the enthalpy of hydration of ammonium chloride (NH4Cl) is approximately -14609 J/mol.
To calculate the enthalpy of hydration of ammonium chloride (NH4Cl) in joules per mole (J/mol), we need to use the equation:
ΔH = q/n
where:
ΔH is the enthalpy change (in J/mol)
q is the heat exchanged (in J)
n is the number of moles
First, we need to determine the heat exchanged during the dissolution process. This can be calculated using the equation:
q = m × c × ΔT
where:
q is the heat exchanged (in J)
m is the mass of the solution (in grams)
c is the specific heat capacity of water (4.18 J/g·°C)
ΔT is the change in temperature (in °C)
Given:
m = 10.5 g (mass of NH4Cl)
c = 4.18 J/g·°C (specific heat capacity of water)
ΔT = (final temperature - initial temperature) = (5.1°C - 22.5°C) = -17.4°C
Now, let's calculate the heat exchanged:
q = 10.5 g × 4.18 J/g·°C × (-17.4°C)
q = -760.23 J (to maintain proper sign conventions, negative value indicated)
Next, we need to determine the number of moles of NH4Cl dissolved. This can be calculated using the equation:
n = m/M
where:
n is the number of moles
m is the mass of NH4Cl (in grams)
M is the molar mass of NH4Cl (in g/mol)
The molar mass of NH4Cl (NH4 = 14.01 g/mol, Cl = 35.45 g/mol) = 14.01 + 4(1.01) + 35.45 = 53.49 g/mol
n = 10.5 g / 53.49 g/mol
n ≈ 0.196 mol
Finally, let's calculate the enthalpy of hydration of ammonium chloride (NH4Cl):
ΔH = q/n
ΔH = -760.23 J / 0.196 mol
ΔH ≈ -3877.2 J/mol
Hence, the enthalpy of hydration of ammonium chloride is approximately -3877.2 J/mol.
q in joules = mass H2O x specific heat x (Tfinal=Tinitial)
That gives you J in 10.5 grams. Convert the 10.5 g to mols and that will be J/? moles. Convert to 1 mols.