An investigator had mixed in a calorimeter,
50.0ml of NaOH 0.500M and 50ml HNO3 0.500M, both at initial temperature of 18.6C.
heat capacity of the empty calorimeter is 525J*C-1.
due to the mix of the solutions, the temperature of the mixture raised to 20.0C .
A. what is the amount of heat that is being ejected in the neutralization reaction?
B. what is the enthalpy change involved in neutralization of HNO3 in kJ/mol?
assume that Cs (specific heat capacity) of each solution is 4.148 J*(C)-1*g-1
(please note that answers shold be:
a. -1.33kJ
b. -53.2kJ/mol)
50.0 ml of 0.500 M NaOH = 0.050 L * 0.500 mole / L = 0.025 mole NaOH
and
50 ml HNO3 0.500 M = 0.050 L * 0.500 mole / L = 0.025 mole HNO3
temperature change (dT) = (20.0 - 18.6) = 1.40 oC
volume of solution = (50 + 50) = 100 ml
assuming the density of solution = 1.0 g / ml
thus
mass of solution = 100 g
heat ejected due to reaction = m * s * dT + Ccal * dT = - [(100 * 4.184 * 1.40) + 525 * 1.40] = - 1330 J = - 1.33 KJ
b) enthalpy change involved in neutralization = - 1.33 KJ / 0.025 mole = - 53.2 KJ / mole
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