A coffee cup calorimeter having a heat capacity of 451 J/degrees Celsius was use

Question

A coffee cup calorimeter having a heat capacity of 451 J/degrees Celsius was used to measure the heat evolved when 0.0300 mole of NaOH(s) was added to 1000 mL of 0.0300 M HNO3 initially at 23.000 degrees Celsius. The temperature of the water rose to 23.639 degrees Celsius. Calculate H (in kJ/mol NaNO3) for this reaction. Assume the specific heat of the final solution is 4.18 J/g degrees Celsius; the density of each solution is 1.00g/ML; and the addition of solid does not appreciably affect the volume of the solution.

HNO3 (aq) + NaOH (s) --> NaNO3 (aq) + H2O (l)

Explanation / Answer

moles of NaOH = 0.0300 mol

moles of HNO3 = 1000 x 0.0300 / 1000 = 0.0300 mol

mass of solution = 1000 g

Cp = 4.18 J/g oC

Q = m Cp dT + Cp dT

    = 1000 x 4.18 x (23.639 - 23) + 451 x (23.639 - 23)

    = 2959.2 J

delta H = - Q / n

            = - 2959.2 x 10^-3 / 0.03

delta H = - 98.6 kJ/mol

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