A metal sample weighting 147.90 g and at a temperature of 99.5 degree C was plac
ID: 896307 • Letter: A
Question
A metal sample weighting 147.90 g and at a temperature of 99.5 degree C was placed in 49.73 g of water in a calorimeter at 23.0 degree C. At equilibrium the temperature of the water and metal was 41.8 degree C. What was Delta t for the metal? How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g degree C.) Calculate the specific heat of the metal, using Equation 3. What is the approximate molar mass of the metal? (Use Eq. 4.) When 4.98 g of NaOH was dissolved in 49.72 g of water in a calorimeter at 23.7 degree C, the temperature of the solution went up to 50.1 degree C. Is this dissolution reaction exothermic? Why? Calculate q_H_2O, using Equation 1. Find Delta H for the reaction as it occurred in the calorimeter (Eq. 5). Find Delta H for the dissolution of 1.00 g NaOH in water. Find Delta H for the dissolution of 1 mole NaOH in water. 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. Given the following heats of formation, Delta H_f, in kJ per mole, as obtained from a table of Delta H_f data, calculate 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), -230.0Explanation / Answer
1. Given data
a. delta T of water = 41.8 - 23.0 = 18.8 oC
b. delta T of metal = 99.5 - 41.8 = 57.7 oC
c. Heat flowed through water q = mCpdT
= 49.73 x 4.18 x 18.8 = 3908.0 J
d. specific heat of metal Cp,
qmetal = qwater
[mCpdT]metal = [mCpdT]water
147.90 x Cp x 57.7 = 3908
Cp = 0.458 J/g.oC
e. molar mass of metal = 25/0.458 = 54.59 g/mol
2. Given data,
a. Reaction is exothermic. NaOH bond is broken when it dissolved in water. This reaction is endothermic. Na+ and OH- ions then form hydrated ions, so it forms two bonds releasing energy during the process. This energy released is greater then the energy needed to break one NaOH bond. So overall the reaction is exothermic.
b. qwater = (49.8 + 49.72) x 4.18 x (50.1 - 23.7) = 10982.23 J
c. delta H = qwater = 10982.23 J
d. delta H/g = 10982.23/49.8 = 220.53 J/g = 0.22 kJ/g
e. Delta H/mol = 220.53 x 40 = 8.82 kJ/mol
f. Na+ + OH- + H2O ---> [Na+]nH2O + [OH-]nH2O
g. Delta H = delta H[products] - delta H[reactants]
= (-240.1 - 230) + 425.6
= -44.5 kJ/mol
So the value for delta H obtained in actual reaction is lower then theoretical value calculated.