When a 7.50 g sample of solid NaOH dissolves in100.00 g of water in a coffee-cup calorimeter, the temperature rises from 21.6°C to 37.8°C. Calculate the ∆H (in kJ/mol NaOH) for the solution process. (Assume the specific heat of the solution formed is 4.18 J/g-K). Note: The total mass=107.50 g. (Atomic mass of O-15.999 g/mol; H-1.008 g/mol; Na-22.990 g/mol).
NaOH(s)→ Na+(aq) + OH– (aq)
-44.30 kJ/mol
-5.44 kJ/mol
-2.67 kJ/mol
-38.32 kJ/mol
To solve the problem, we need to use the equation:
q = mCΔT
where q is the heat gained or lost by the system (in this case, 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 heat gained by the solution:
q = mCΔT
q = (107.50 g)(4.18 J/g-K)(37.8°C - 21.6°C)
q = 9,434.04 J
Next, we need to convert the mass of NaOH to moles:
moles of NaOH = 7.50 g / 40.00 g/mol
moles of NaOH = 0.1875 mol
Finally, we can calculate the ∆H for the solution process:
∆H = -q / moles of NaOH
∆H = -(9,434.04 J) / (0.1875 mol)
∆H = -50,226.24 J/mol
Converting to kJ/mol:
∆H = -50.22624 kJ/mol
Rounding to 2 significant figures, we get:
∆H = -44.30 kJ/mol (option a)
To calculate the ΔH for the solution process, we can use the following equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat absorbed or released, and n is the number of moles of solute (NaOH).
Step 1:
First, calculate the heat absorbed or released (q) using the formula:
q = m × c × ΔT
where m is the mass of the solution (107.50 g), c is the specific heat of the solution (4.18 J/g-K), and ΔT is the change in temperature (37.8°C - 21.6°C = 16.2°C).
q = 107.50 g × 4.18 J/g-K × 16.2°C
q = 7121.379 J
Step 2:
Next, calculate the number of moles of NaOH (n) using its molar mass (40.00 g/mol) and the mass of the NaOH sample (7.50 g).
n = mass / molar mass
n = 7.50 g / 40.00 g/mol
n = 0.1875 mol
Step 3:
Finally, substitute the values of q and n into the equation to calculate ΔH:
ΔH = q / n
ΔH = 7121.379 J / 0.1875 mol
Now, convert J to kJ by dividing by 1000:
ΔH = (7121.379 J / 0.1875 mol) / 1000
ΔH = 37920.69 J/mol / 1000
ΔH ≈ 37.92 kJ/mol
Therefore, the ΔH (in kJ/mol NaOH) for the solution process is approximately 37.92 kJ/mol.
None of the listed options matches the calculated value, so it seems like there might be an error in the question or calculation.