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)
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
1 answer
1 answer