10.8 g of solid ammonium nitrate (NH4NO3, molar mass 80.052 g/mol) is dissolved in 150 mL of water.
The temperature change, ΔT, of the system is -5.1 ºC.
Calculate the heat of solution for ammonium nitrate in kJ/mol.
i did H=mct as .225 but thats wrong
I assume you used 10.8 g for mass but the water is what's cooled by the NH4NO3 so the water gives you the amount of cooling and that's for 10.8 g NH4NO3.
q = m*c*dT
Assuming density of water is 1 g/mL, mass H2O is 150 g.
q = 150 g x 4.184 J/g*C x -5.1 C = -3,201 kJ for 10.8 g NH4NO3.
mols NH4NO3 = 10.8 g/80.052 = 0.134 mols.
So delta H is 3201 kJ/0.135 mols or 23,711 kJ/mol NH4NO3.
You should go through and check my math. I estimated here and there and didn't use all of the decimal places.
To calculate the heat of solution for ammonium nitrate, we can use the equation:
q = m × c × ΔT
where:
- q is the heat absorbed or released by the system,
- m is the mass of the ammonium nitrate dissolved in grams,
- c is the specific heat capacity of water (4.18 J/g·°C),
- ΔT is the change in temperature in degrees Celsius.
First, we need to convert the mass of the ammonium nitrate from grams to moles. We can use the molar mass of ammonium nitrate to do this:
moles = mass / molar mass
moles = 10.8 g / 80.052 g/mol
moles = 0.135 mol
Next, we can calculate the heat absorbed or released by the system using the equation:
q = m × c × ΔT
q = 0.135 mol × 4.18 J/g·°C × (-5.1 ºC)
Now, let's perform the calculation:
q = 0.135 mol × 4.18 J/g·°C × (-5.1 ºC)
q = -2.85303 J
Finally, we need to convert the heat from joules to kilojoules:
q(kJ) = q(J) / 1000
q(kJ) = -2.85303 J / 1000
q(kJ) = -0.00285303 kJ
Therefore, the heat of solution for ammonium nitrate is approximately -0.002853 kJ/mol.