when 2g of NaOH were dissolved in 149 g of H2o in a calorimeter at 24 degrees C the temperature of the solution went up to 29 degrees C. If the calorimeter constant was 11 cal/C find delta H for the solution of one mole of NaOH in H2O
find the deltaH per gram NaOH.
it must be 1/2 ((149*cwater*5)+11*5)
now multiply by the mol mass of NaOH (40).
this does not help . . . . can any one else help me please
q = heat to water + heat to calorimeter
q = [(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal*(Tfinal-Tinital)]
Then dH in cal/g= q/2g
and dH in cal/mol = cal/g x molar mass NaOH
To find the enthalpy change (ΔH) for the solution of one mole of NaOH in H2O, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat absorbed or released by the reaction, and n is the number of moles of the substance involved.
In this case, we have the following information:
- Mass of NaOH: 2g
- Mass of H2O: 149g
- Change in temperature (ΔT): 29°C - 24°C = 5°C
- Calorimeter constant: 11 cal/°C
First, let's find the heat absorbed or released by the reaction (q).
Since we have the change in temperature and the calorimeter constant, we can use the formula:
q = Ccal × ΔT
where Ccal is the calorimeter constant and ΔT is the change in temperature.
Plugging in the values:
q = 11 cal/°C × 5°C = 55 cal
Next, we need to calculate the number of moles of NaOH.
To do this, we'll use the molecular weight (MW) of NaOH, which is 23 g/mol for Na + 16 g/mol for O + 1 g/mol for H:
MW of NaOH = 23 g/mol + 16 g/mol + 1 g/mol = 40 g/mol
Since we have 2g of NaOH, the number of moles (n) can be calculated using the formula:
n = mass / MW
n = 2g / 40 g/mol = 0.05 mol
Finally, we can find the enthalpy change (ΔH) using the formula mentioned earlier:
ΔH = q / n
Plugging in the values:
ΔH = 55 cal / 0.05 mol = 1100 cal/mol
Therefore, the enthalpy change for the solution of one mole of NaOH in H2O is 1100 cal/mol.