When sulfuric acid dissolves in water, a great deal of heat is given off. The enthalpy change for this process is called the enthalpy of solution. To measure it, 175 mL of water was placed in a coffee-cup calorimeter and chilled to 10oC. Then 49.0 mL of pure sulfuric acid, also at 10.0oC was added, and the mixture was quickly stirred with a thermometer. The temperature rose rapidly to 14.9oC Pure sulfuric acid has a concentration of 17.8 mol/L Calculate the molar enthalpy of this dilution kJ/mol
To calculate the molar enthalpy of this dilution, we can use the equation:
q = mcΔT
where q is the heat gained or lost by the solution, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, we need to calculate the mass of the solution. The density of water is 1 g/mL, so the mass of 175 mL of water is:
mass of water = density × volume
mass of water = (1 g/mL) × (175 mL)
mass of water = 175 g
Next, we need to calculate the specific heat capacity of the solution. Since the solution is primarily water, we can assume that the specific heat capacity of the solution is the same as that of water, which is 4.18 J/g·°C.
Now, we can calculate the heat gained or lost by the solution, q:
q = (mass of water + mass of sulfuric acid) × c × ΔT
Substituting the given values:
q = (175 g + 49 g) × (4.18 J/g·°C) × (14.9°C - 10.0°C)
q = 224 g × 4.18 J/g·°C × 4.9°C
q = 4645.632 J
Next, we need to convert the heat gained or lost by the solution, q, to kilojoules:
q = 4645.632 J ÷ 1000
q = 4.645632 kJ
Finally, we can calculate the molar enthalpy of the dilution:
moles of sulfuric acid = volume of sulfuric acid × concentration of sulfuric acid
moles of sulfuric acid = 49 mL × (1 L/1000 mL) × 17.8 mol/L
moles of sulfuric acid = 0.8702 mol
molar enthalpy of dilution = heat gained or lost (q) ÷ moles of sulfuric acid
molar enthalpy of dilution = 4.645632 kJ ÷ 0.8702 mol
molar enthalpy of dilution ≈ 5.34 kJ/mol
Therefore, the molar enthalpy of this dilution is approximately 5.34 kJ/mol.
To calculate the molar enthalpy of the dilution, we can use the equation:
q = m * C * ΔT
where:
q is the amount of heat absorbed or released (in joules),
m is the mass of the solution (in grams),
C is the specific heat capacity of the solution (in J/g·°C), and
ΔT is the change in temperature (in °C).
First, let's calculate the mass of the solution. We have two solutions (water and sulfuric acid) that are being mixed together:
Mass of water = volume of water * density of water
= 175 mL * 1 g/mL (since the density of water is approximately 1 g/mL)
= 175 g
Mass of sulfuric acid = volume of sulfuric acid * density of sulfuric acid
= 49.0 mL * 1.84 g/mL (from the provided density of sulfuric acid)
= 90.16 g
Total mass of the solution = mass of water + mass of sulfuric acid
= 175 g + 90.16 g
= 265.16 g
Next, let's calculate the change in temperature:
ΔT = Final temperature - Initial temperature
= 14.9 °C - 10.0 °C
= 4.9 °C
Now, we need to determine the specific heat capacity of the solution. Since we know the specific heat capacity of water, we can assume the specific heat capacity of the solution to be approximately the same:
C = 4.18 J/g·°C (specific heat capacity of water)
Finally, we can calculate the amount of heat released or absorbed (q):
q = m * C * ΔT
= 265.16 g * 4.18 J/g·°C * 4.9 °C
= 5530.07 J
To convert the amount of heat to kilojoules, divide by 1000:
q = 5530.07 J / 1000
= 5.53007 kJ
Lastly, since the enthalpy change for this process is per mole of sulfuric acid, we need to determine the number of moles of sulfuric acid present:
Concentration of sulfuric acid = 17.8 mol/L
Moles of sulfuric acid = volume of sulfuric acid * concentration of sulfuric acid
= 49.0 mL * 0.0178 mol/L
= 0.8722 mol
Finally, we can calculate the molar enthalpy of this dilution:
Molar enthalpy of dilution = q / moles of sulfuric acid
= 5.53007 kJ / 0.8722 mol
≈ 6.34 kJ/mol
Therefore, the molar enthalpy of this dilution is approximately 6.34 kJ/mol.
To calculate the molar enthalpy of the dilution, we can use the equation:
q = mcΔT
Where:
q = heat gained or lost by the solution (in joules)
m = mass of the solution (in grams)
c = specific heat capacity of the solution (in J/g°C)
ΔT = change in temperature (in °C)
First, we need to determine the initial and final temperatures of the solution:
Initial temperature (Ti) = 10°C
Final temperature (Tf) = 14.9°C
Next, we need to calculate the mass of the solution. Since we know the volume of water and sulfuric acid, as well as their densities, we can calculate their masses:
Mass of water = volume of water × density of water
= 175 mL × 1 g/mL
= 175 grams
Mass of sulfuric acid = volume of sulfuric acid × density of sulfuric acid
= 49.0 mL × 1.84 g/mL (density of 17.8 M sulfuric acid)
= 90.16 grams
The total mass of the solution is then the sum of the masses of water and sulfuric acid:
Total mass of the solution = mass of water + mass of sulfuric acid
= 175 grams + 90.16 grams
= 265.16 grams
Next, we need to calculate the change in temperature:
ΔT = Tf - Ti
= 14.9°C - 10°C
= 4.9°C
Now, we need to calculate the heat gained or lost by the solution (q). We can assume that the heat lost by the sulfuric acid is equal to the heat gained by the water:
q = -q(sulfuric acid)
= q(water)
= mcΔT
We know the specific heat capacity (c) of water is 4.18 J/g°C.
q = mcΔT
= (mass of water) × (specific heat capacity of water) × (change in temperature)
= 175 g × 4.18 J/g°C × 4.9°C
= 3410.15 J
Next, we need to convert the energy from joules to kilojoules:
1 kJ = 1000 J
Therefore, the energy in kilojoules is:
Energy (q) = 3410.15 J ÷ 1000
= 3.41015 kJ
Finally, we can calculate the molar enthalpy of the dilution using the equation:
Molar enthalpy (ΔH) = Energy (q) ÷ moles of sulfuric acid
To find the moles of sulfuric acid, we can use the concentration (C) and the volume (V) of the sulfuric acid:
moles of sulfuric acid = concentration × volume of sulfuric acid
= 17.8 mol/L × 0.049 L (volume in liters)
= 0.8712 mol
Therefore, the molar enthalpy of the dilution is:
Molar enthalpy (ΔH) = 3.41015 kJ ÷ 0.8712 mol
≈ 3.918 kJ/mol
So, the molar enthalpy of the dilution is approximately 3.918 kJ/mol.