A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves 25.0 mL of water also at 25.0 degrees celsius. The final equilibrium temperature of the resulting soltution is 18.1 degrees celsius. What is the enthalpy of of solution in kilojoules per mole of KBr?
To find the enthalpy of solution for KBr, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy of solution, q is the heat exchanged (in joules), and n is the number of moles of solute (KBr) dissolved.
The heat exchanged, q, can be calculated using the formula:
q = m * c * ΔT
where m is the mass of the water (in grams), c is the specific heat capacity of water (4.18 J/g·°C), and ΔT is the change in temperature (in °C).
First, we need to calculate the mass of water:
density of water = 1 g/mL
mass of water = volume of water * density of water
= 25.0 mL * 1 g/mL
= 25.0 g
Next, we calculate the change in temperature:
ΔT = final temperature - initial temperature
= 18.1°C - 25.0°C
= -6.9°C
Next, we can calculate the heat exchanged:
q = m * c * ΔT
= 25.0 g * 4.18 J/g·°C * (-6.9°C)
= -717.15 J
Since we are given the mass of KBr (5.0 g), we can calculate the number of moles of KBr using its molar mass.
Molar mass of KBr = 39.10 g/mol + 79.90 g/mol
= 119.00 g/mol
n = mass of KBr / molar mass of KBr
= 5.0 g / 119.00 g/mol
= 0.042 mol
Finally, we can calculate the enthalpy of solution:
ΔH = q / n
= -717.15 J / 0.042 mol
≈ -17,079 J/mol
To convert the result to kilojoules per mole, we divide by 1000:
ΔH ≈ -17.08 kJ/mol
Therefore, the enthalpy of solution for KBr is approximately -17.08 kJ/mol.
To calculate the enthalpy of solution (ΔH_solution) in kilojoules per mole of KBr, we can use the equation:
ΔH_solution = q_solution / n_KBr
Where:
q_solution = heat exchanged during the solution process
n_KBr = number of moles of KBr dissolved
First, let's calculate the heat exchanged (q_solution) during the solution process using the formula:
q_solution = m_solution x C_solution x ΔT_solution
Where:
m_solution = mass of water in grams
C_solution = specific heat capacity of water in J/(g°C)
ΔT_solution = change in temperature of the solution in °C
Given:
Mass of water (m_solution) = 25.0 mL = 25.0 g
Specific heat capacity of water (C_solution) = 4.18 J/(g°C)
Change in temperature of the solution (ΔT_solution) = 25.0 - 18.1 = 6.9 °C
Plugging in the values:
q_solution = 25.0 g x 4.18 J/(g°C) x 6.9 °C
q_solution = 717.675 J
Next, let's calculate the number of moles of KBr dissolved (n_KBr) using the formula:
n_KBr = mass_KBr / molar mass_KBr
Given:
Mass of KBr (mass_KBr) = 5.0 g
Molar mass of KBr (molar mass_KBr) = 119.0 g/mol
Plugging in the values:
n_KBr = 5.0 g / 119.0 g/mol
n_KBr = 0.0420 mol
Now, we can substitute the values into the formula to calculate the enthalpy of solution (ΔH_solution):
ΔH_solution = q_solution / n_KBr
ΔH_solution = 717.675 J / 0.0420 mol
ΔH_solution = 17,086.607 J/mol
Finally, let's convert the value to kilojoules per mole by dividing by 1000:
ΔH_solution = 17,086.607 J/mol / 1000
ΔH_solution ≈ 17.1 kJ/mol
Therefore, the enthalpy of solution of KBr is approximately 17.1 kJ/mol.
q H2O = mass H2O x specific heat H2O x (Tfinal-Tinitial)
delta H/gram = q/5.0g
delta H/mol = (q/5.0g) x (molar mass KBr/mol) = ?