In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g
water at an initial temperature of 23.50 ºC. When the salt was
completely dissolved, the solution temperature of the calorimeter
became 21.80 ºC. Assume the solution’s specific heat capacity is
4.184 J/g. ºC.
Calculate the enthalpy change (ΔH) for the
dissolution of NH4NO3 in J/g and J/mol
i want answer
Yes
To calculate the enthalpy change (ΔH) for the dissolution of NH4NO3 in J/g and J/mol, we need to use the equation:
ΔH = q / m
where ΔH is the enthalpy change, q is the heat absorbed or released during the process, and m is the mass of the substance.
First, let's calculate the heat absorbed or released during the process using the equation:
q = m * C * ΔT
where q is the heat absorbed or released, m is the mass of the water, C is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of NH4NO3 = 1.60 g
Mass of water = 75.0 g
Initial temperature = 23.50 ºC
Final temperature = 21.80 ºC
Specific heat capacity (C) = 4.184 J/g.ºC
To calculate the heat absorbed or released by the water, we need to use the equation:
q = m * C * ΔT
ΔT = final temperature - initial temperature
= 21.80 ºC - 23.50 ºC
= -1.70 ºC
q = 75.0 g * 4.184 J/g.ºC * (-1.70 ºC)
= -540.60 J
The negative sign indicates that heat is released by the water during the process.
Now, let's calculate the enthalpy change (ΔH) for the dissolution of NH4NO3 using the equation:
ΔH = q / m
ΔH = -540.60 J / 1.60 g
= -338.0 J/g
To calculate the enthalpy change in J/mol, we need to convert the mass from g to mol using the molar mass of NH4NO3.
Molar mass of NH4NO3 = 14.01 g/mol (N) + 1.01 g/mol (H) + 14.01 g/mol (N) + 3(16.00 g/mol) (O)
= 80.04 g/mol
Now we can calculate the enthalpy change (ΔH) in J/mol.
ΔH = ΔH (J/g) * molar mass (g/mol)
ΔH = -338.0 J/g * 80.04 g/mol
= -27053.52 J/mol
Therefore, the enthalpy change (ΔH) for the dissolution of NH4NO3 is -338.0 J/g and -27053.52 J/mol.