When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5 C,

Question

When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5 C, the temperature of the solution went up to 49.8 C.

a. Calculate qH20.
b. Find delta H for the reaction as it occurred in the calorimeter.
c. Find delta H for the solution of 1.00 g NaOH in water.
d. Find delta H for the solution of 1 mole NaOH in water.
f. Given that NaOh exists as Na+ and OH_ ions in solution, write the equation for the reaction that occurs when NaOH is dissolved in water.
g. Given the following heats of formation, delta H f, in kJ per mole, as obtained from a table of deltaHf data, calculate the delta H for the reaction in part f. Compare your answer with the result you obtained in Part e.
NaOH(s),-425.6; Na+(aq),-240.1;OH-(aq),-230.0

Thanks in advance!

Explanation / Answer

a.) qH2O = mcT = 51.55g * 4.184J/gC * (49.8-24.5C) = 5457 J

b.) H = -5457 Joules

c.) H/g = -5457 Joules/5.12 g = -1066 J/g

d.) H/mole = -1066J/g * 40g/mole = -42,600 J/mole = -42.6 kJ/mole

f.) NaOH(s) ===> Na+(aq) + OH-(aq)

g.) H = nHf(products) -nHf(reactants)

H = (-230 - 240.1) - (-425.6)

H = -44.5 kJ/mole

The values are very close, so our experiment did a fairly good job of measuring the enthalpy.

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