When 0.917 g of CaO is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 3.76C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
Hrxn, for the reaction of
CaO(s) + 2H+(aq) Ca2+(aq) + H2O(l)
is _________ kJ/mol.
-192 because .....
- delta H = [(200)(4.184)(3.76)(1/1000)]/ [(.917/56.077)]
what is the 56.077 at the last part of this question?
It's the molar mass for CaO
To find the enthalpy change (ΔHrxn) for the given reaction, we can use the equation:
ΔHrxn = q / n
where q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant.
First, let's find the heat change (q) using the equation:
q = mcΔT
where m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
Given:
Mass of solution (m) = 200.0 mL = 200.0 g
Specific heat capacity (c) = 4.184 J/g°C
Change in temperature (ΔT) = 3.76°C
Substituting these values into the equation:
q = (200.0 g) * (4.184 J/g°C) * (3.76°C)
q = 3151.9 J
Next, let's calculate the number of moles of the limiting reactant (CaO) used:
Molar mass of CaO = 40.08 g/mol
Mass of CaO used = 0.917 g
Using the equation:
moles = mass / molar mass
moles of CaO = 0.917 g / 40.08 g/mol
moles of CaO ≈ 0.02288 mol
Now we can substitute the values into the formula for ΔHrxn:
ΔHrxn = q / n
ΔHrxn = 3151.9 J / 0.02288 mol
Calculating this value gives us:
ΔHrxn ≈ -137,576 J/mol
Converting this to kilojoules:
ΔHrxn ≈ -137.576 kJ/mol
Rounding to the appropriate number of significant figures, the enthalpy change (ΔHrxn) for the reaction of CaO(s) + 2H+(aq) → Ca2+(aq) + H2O(l) is approximately -137.6 kJ/mol.