A 6.3 g sample of solid NH4NO3 is dissolved in 50g of water, causing the temperature to drop from 24°C to 19°C. Calculate the ∆Hsolution in J/mol.
-Thank you ☺
The answer will be in J/grams. Convert to J/mol
See your post for q below. In this case, q is the same as dHsolution.
To calculate the ∆Hsolution in J/mol, we can use the equation:
∆Hsolution = q / n
where q is the heat released or absorbed by the solution, and n is the number of moles of NH4NO3.
First, let's find the heat released or absorbed by the solution (q). We can use the equation:
q = m × C × ∆T
where m is the mass of water, C is the specific heat capacity of water, and ∆T is the change in temperature.
Given:
Mass of water (m) = 50g
Change in temperature (∆T) = 19°C - 24°C = -5°C
The specific heat capacity of water (C) is 4.18 J/g·°C.
q = 50g × 4.18 J/g·°C × -5°C
q = -1045 J
Now, let's calculate the number of moles of NH4NO3 (n). We can use the equation:
n = mass / molar mass
Given:
Mass of NH4NO3 = 6.3g
Molar mass of NH4NO3 = 80.0434 g/mol (calculated from the atomic masses of nitrogen (N), hydrogen (H), and oxygen (O))
n = 6.3g / 80.0434 g/mol
n = 0.0787 mol
Finally, we can calculate the ∆Hsolution:
∆Hsolution = q / n
∆Hsolution = -1045 J / 0.0787 mol
∆Hsolution ≈ -13277 J/mol
Therefore, the ∆Hsolution for the dissolution of NH4NO3 is approximately -13277 J/mol.
To calculate the ∆Hsolution, we can use the equation:
∆Hsolution = q / n
Where:
∆Hsolution is the enthalpy change for the solution (in J/mol)
q is the amount of heat released or absorbed by the solution (in J)
n is the number of moles of solute (NH4NO3)
To find the value of q, we can use the equation:
q = m * c * ∆T
Where:
m is the mass of the solution (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
∆T is the change in temperature (in °C)
First, calculate the mass of the solution:
Mass of the solution = mass of water + mass of solute
Mass of the solution = 50 g + 6.3 g
Mass of the solution = 56.3 g
Next, calculate the change in temperature:
∆T = Final temperature - Initial temperature
∆T = 19°C - 24°C
∆T = -5°C
Now, calculate the value of q using the equation:
q = m * c * ∆T
q = 56.3 g * 4.18 J/g°C * -5°C
q = -1178.45 J
Lastly, calculate the number of moles of NH4NO3:
Number of moles = mass / molar mass
Number of moles = 6.3 g / (14.01 g/mol + 14.01 g/mol + 14.01 g/mol + 16.00 g/mol)
Number of moles = 6.3 g / 80.04 g/mol
Number of moles = 0.0787 mol
Now, we can use the equation ∆Hsolution = q / n to calculate the enthalpy change for the solution:
∆Hsolution = -1178.45 J / 0.0787 mol
∆Hsolution ≈ -14,945 J/mol
Therefore, the enthalpy change for the solution (∆Hsolution) is approximately -14,945 J/mol.