Question

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 C_{s (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)}

Answer 1

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