Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
Answer:
-55.8 kJ/mol
Explanation:
There is a part missing from the question.
Assume no heat is lost to the calorimeter or the surroundings, and the density and heat capacity of the resulting solution are the same as water.
The initial moles of NaOH and HNO₃ are:
0.1000 L × 0.300 mol/L = 3.00 × 10⁻² mol
The neutralization reaction is:
NaOH + HNO₃ → NaNO₃ + H₂O
When 3.00 × 10⁻² moles of NaOH react with 3.00 × 10⁻² moles of HNO₃, they produce 3.00 × 10⁻² moles of NaNO₃ and 3.00 × 10⁻² moles of H₂O.
According to the law of conservation of energy, the sum of the heat released by the reaction and the heat absorbed by the solution is equal to zero.
ΔH°rxn + ΔH°sol = 0
ΔH°rxn = -ΔH°sol [1]
The volume of the solution is 100.0 mL + 100.0 mL = 200.0 mL. Since the density is 1.00 g/mL, the mass of the solution is 200.0 g.
We can calculate the heat absorbed by the solution using the following expression.
ΔH°sol = c × m × ΔT = (4.184 × 10⁻³ kJ/g.°C) × 200.0 g × (37.00°C - 35.00°C) = 1.674 kJ
where,
c: specific heat capacity of the solution
m: mass of the solution
ΔT: change in the temperature
From [1],
ΔH°rxn = -1.674 kJ
We can express the enthalpy of reaction per mole of NaNO₃.
ΔH°rxn = -1.674 kJ / 3.00 × 10⁻² mol = -55.8 kJ/mol
