the temperature of 50.0g of water changes from 28.5°C to 26.9°C when 1.00g of ammonium bromide, NH4br(s), dissolves. What is the molar enthalpy of solution for this compound?
The change of T of water is the way you get to the dH soln.
q = mass H2O x specific heat H2O x (28.5-26.9) = ? in joules if sp.j. in joules.
q/50.0 g = delta H/gram
(delta H/gram) x (molar mass NH4Br) = delta H in J/mol. The usual way of expressing this is in kJ/mol convert J to kJ for this.
To calculate the molar enthalpy of solution for ammonium bromide (NH4Br), we need to use the formula:
ΔH = q / n
Where:
ΔH = Molar enthalpy of solution
q = Heat absorbed or released during the process (in Joules)
n = Number of moles of the solute (NH4Br)
First, we need to find the heat absorbed or released during the process (q). We can use the equation:
q = m × c × ΔT
Where:
q = Heat absorbed or released (in Joules)
m = Mass of the water (50.0g)
c = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature (final temperature - initial temperature)
ΔT = 26.9°C - 28.5°C
ΔT = -1.6°C
Now let's convert the mass of NH4Br to moles.
molar mass of NH4Br = molar mass of N + 4 × molar mass of H + molar mass of Br
molar mass of NH4Br = 14.01 g/mol + 4 × 1.01 g/mol + 79.90 g/mol
molar mass of NH4Br = 97.04 g/mol
moles of NH4Br = mass of NH4Br / molar mass of NH4Br
moles of NH4Br = 1.00g / 97.04 g/mol
Now we can plug the values into the first equation to find the molar enthalpy of solution.
ΔH = q / n
ΔH = (m × c × ΔT) / (moles of NH4Br)
To calculate the molar enthalpy of solution for ammonium bromide (NH4Br), we need to use the equation:
ΔHsolution = Q / n
Where:
ΔHsolution is the molar enthalpy of solution
Q is the heat gained or lost during the process (in joules)
n is the number of moles of the solute
To find the value of Q, we need to use the equation:
Q = mcΔT
Where:
m is the mass of the water (in grams)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT is the change in temperature (in Celsius)
First, let's calculate Q:
m = 50.0 g (mass of water)
c = 4.184 J/g°C (specific heat capacity of water)
ΔT = 26.9°C - 28.5°C = -1.6°C
Q = (50.0 g) * (4.184 J/g°C) * (-1.6°C)
Q = -334.4 J
Next, we need to calculate the number of moles of NH4Br dissolved. To do this, we can use the molar mass of NH4Br:
Molar mass of NH4Br = (4 * 1.01 g/mol) + (1 * 14.01 g/mol) + (1 * 79.90 g/mol)
Molar mass of NH4Br = 97.05 g/mol
n = mass / molar mass
n = 1.00 g / 97.05 g/mol
n = 0.0103 mol
Now we can calculate the molar enthalpy of solution (ΔHsolution):
ΔHsolution = Q / n
ΔHsolution = -334.4 J / 0.0103 mol
ΔHsolution ≈ -32,504 J/mol
Therefore, the molar enthalpy of solution for ammonium bromide (NH4Br) is approximately -32,504 J/mol.