A 5.0g KBr sample, at 25 degrees celsius, dissolved in 25 ml of water, also at 25 degrees celsius. the final equilibrium temperature of the resulting solution is 18.1 degrees celsius, What is the enthalpy change in J/g and KJ/mole of KBr.
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)]
Substitute and solve for q = delta H for the solution.
q/5 = delta H/g in joules.
delta (H/g) x molar mass KBr = J/mol.
Convert to kJ/mol.
To calculate the enthalpy change (∆H) in J/g and kJ/mol of KBr, you need to use the equation:
∆H = q / m
where:
∆H is the enthalpy change,
q is the heat exchanged,
m is the mass of the substance.
First, let's find the heat exchanged (q).
To do that, we can use the equation:
q = m × c × ∆T
where:
q is the heat exchanged,
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.18 J/g°C), and
∆T is the change in temperature of the water.
In this case, the mass of the water is 25 mL, but we need it in grams. Since the density of water is approximately 1 g/mL, the mass of the water is 25 g.
∆T represents the change in temperature, which is calculated by subtracting the final temperature (18.1°C) from the initial temperature (25°C). Therefore, ∆T = 25°C - 18.1°C = 6.9°C.
Now we can calculate q:
q = (25 g) × (4.18 J/g°C) × (6.9°C) = 727.725 J
Next, we need to calculate the mass (m) of KBr. Given that the sample weighs 5.0 g, we have m = 5.0 g.
Finally, we can calculate the enthalpy change (∆H).
∆H = q / m = (727.725 J) / (5.0 g) = 145.545 J/g
To convert this to kJ/mol, we need to calculate the molar mass of KBr. The molar mass of KBr is 119 g/mol (39 g/mol for potassium (K) and 80 g/mol for bromine (Br)).
Now we can calculate the enthalpy change (∆H) in kJ/mol:
∆H = (145.545 J/g) × (1 kJ / 1000 J) × (1 mol / 119 g) = 0.00122096 kJ/mol
Therefore, the enthalpy change of the KBr sample is approximately 0.145 J/g and 0.00122 kJ/mol.