Julie is conducting an experiment where she placed 29.5 mL of water in a calorimeter at 15.0 °C. Then, 2.4 g of A (molar mass = 48.0 g/mol), also at 15.0°C, is added to the water in the calorimeter and the temperature of the solution increases to 28.0 °C. The following reaction is produced: A(s)--->A(aq)

Question

Julie is conducting an experiment where she placed 29.5 mL of water in a calorimeter at 15.0 °C. Then, 2.4 g of A (molar mass = 48.0 g/mol), also at 15.0°C, is added to the water in the calorimeter and the temperature of the solution increases to 28.0 °C. The following reaction is produced: A(s)--->A(aq)

a.) What is the enthalpy of the reaction (ΔH°rxn), in kJ/mol?

Answer 1

heat added: water heat

= 29.5*cw*(28-15)

moles A=2.4/48

enthalpy= heatadded/molesA

Answer 2

Question

If there is a Part B.
Like Using the following information:
B(s)--B(aq) Delta H=+23.8 KJ/mole
2B(aq)---A (aq) Delta H= -1.25 KJ/mole
Calculate ÄH°rxn for the following reaction (in kJ/mol).
A(s)---2B(s)
How exactly would you do that?

Answer 3

To find the enthalpy of the reaction (ΔH°rxn), we can use the equation:

ΔH°rxn = q / n

where:
ΔH°rxn is the enthalpy change of the reaction (in kJ/mol)
q is the heat absorbed or released by the reaction (in kJ)
n is the amount of substance involved in the reaction in moles

To calculate the heat absorbed or released by the reaction (q), we can use the equation:

q = mcΔT

where:
q is the heat absorbed or released by the reaction (in kJ)
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)

1. First, we need to calculate the mass of the solution.
The mass of the solution is the sum of the mass of water and the mass of substance A:

mass of water = volume of water * density of water
mass of water = 29.5 mL * 1 g/mL = 29.5 g

mass of substance A = 2.4 g

mass of solution = mass of water + mass of substance A
mass of solution = 29.5 g + 2.4 g = 31.9 g

2. Next, we can calculate the change in temperature (ΔT):

ΔT = final temperature - initial temperature
ΔT = 28.0 °C - 15.0 °C = 13.0 °C

3. Now, we can calculate the heat absorbed or released by the reaction (q):

q = mcΔT
q = (31.9 g) * (4.18 J/g°C) * (13.0 °C)
q = 1739.214 J

(Note: The heat obtained is in joules, but we will convert it to kJ later.)

4. We need to convert the heat obtained (q) to kJ:

1 kJ = 1000 J
q (in kJ) = 1739.214 J / 1000 = 1.739214 kJ

5. Finally, we can calculate the enthalpy change of the reaction (ΔH°rxn):

ΔH°rxn = q / n
Since the molar mass of substance A is given, we can calculate the number of moles (n) of A:

n = mass / molar mass
n = 2.4 g / 48.0 g/mol = 0.05 mol

ΔH°rxn = 1.739214 kJ / 0.05 mol
ΔH°rxn ≈ 34.784 kJ/mol

Therefore, the enthalpy of the reaction (ΔH°rxn) is approximately 34.784 kJ/mol.