HELP with chemistry!?
I'm so lost with this problem
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.
- siggyLv 69 years agoFavorite Answer
Formula is -
NaOH + HNO3 => H2O + NaNO3
moles of base = moles acid = CV = 0.300 x 0.100 = 0.03 moles
Heat produced = ∆ H = mC∆T = 200 x 4.18 x 2 = 1672 Joules
where m = mass of water, C = heat capacity water = 4.18 J/g/K
∆ T = temperature change of water
Since you get 1672 J per 0.03 moles reactant, then -
the heat evolution for 1 mole would be 1672 x 1 / 0.03 = 55730 J = 55.73 kj/mol
That's close - the actual figure should be 57.1 kJ/mol. .Source(s): chem background