Name Section Experiment 14 Advance Study Assignment: Heat Effects and Calorimetr
ID: 1061191 • Letter: N
Question
Name Section Experiment 14 Advance Study Assignment: Heat Effects and Calorimetry i. A metal sample weighing 124 10 g and at a temperature of 99.3 C was placed in 42.87 g of water in a calorimeter at 239 C At equilibrium the temperature of the water and metal was 41.6 C. a What was A for the water? (Ar b What was Au for the metal? c, How much beat flowed into the water? joules d Taking the specific heat of water to be 4,18 Jg C, calculate the specific heat of the metal. using Equation 3. joulesg C e, What is the approximate molar mass of the metal? Use Eq. 4.) 2. When s 12 g of NaoH were dissolved in 5155 g water in a calorimeter at 24.5 C, the temperature of the solution went up to 498 c si Is this solution reaction exothermic? Why? Calculate using Equation 1. b joules e. Find AM for the reaction as it occurred in the calorimeter (Eq. 5). joules (continued on following page)Explanation / Answer
1a) t for the water (t = tfinal – tinitial) = (41.6 – 23.9)C = 17.7C
b) t for the metal = (99.3 – 41.6)C = 57.7C (Here we did t = tinitial – tfinal since tinitial > tfinal)
c) Heat flowed into the water = (mass of water)*(specific heat capacity of water)*(change in temperature) = (42.87 g)*(4.18 J/g.C)*(17.7C) = 3171.7798 joules 3171.78 joules
d) Specific heat of the metal: We know that heat lost by metal = heat gained by water.
Let the specific heat capacity of the metal be S. Therefore, heat lost by the metal = (mass of metal)*(specific heat capacity of metal)*(change in temperature) = (124.10 g)*(S)*(57.7C) = (7160.57 g.C)*S
As per the principle of thermochemistry,
(7160.57*S)g.C = 3171.78 joules
====> S = 0.4429 J/g.C 0.44 J/g.C (ans).
e) I need equation 4; kindly provide.
2a) The reaction is exothermic because dissolution of NaOH into water lead to generation of heat and rise in temperature of the solution.
b) qH2O = (mass of water)*(specific heat capacity of water)*(final temperature – initial temperature) = (51.55 g)*(4.18 J/g.C)*(49.8 – 24.5)C (the initial temperature of water is the same as the temperature of the calorimeter) = 5451.6187 joules 5451.62 joules
c) H for the reaction: The reaction released heat which led to increase in temperature of the water; hence the heat released by the reaction must be equal in magnitude but opposite in sign to that qH2O.
Therefore, H for the reaction = -5451.62 joules
d) H for the solution of 1.00 g of NaOH = (H for the reaction)/(5.12 g NaOH)*(1.00 g NaOH) = (-5451.62 joules)/(5.12 g)*(1.00 g) = -1064.7695 joules -1064.77 joules.
e) 1 mole NaOH = 40 g/mol NaOH
H for solution of 1 mole NaOH in water = (-1064.77 joules/1.00 g)*(40.0 g/1 mole) = -42590.77 joules/mol = -42.59 kJ/mol (ans).
f) The reaction is
NaOH (s) ------> Na+ (aq) + OH- (aq)
g) H = [Hf (Na+) + Hf (OH-)] – [Hf (NaOH)] = [(-240.1 kJ/mol) + (-230.0 kJ/mol)] – (-425.6 kJ/mol) = (-470.1 kJ/mol) + 425.6 kJ/mol = -44.5 kJ/mol (ans).
The value for H which we obtained = -42.59 kJ/mol and the theoretically determined value is -44.5 kJ/mol. The two values are close to each other and the error in determining the experimental value is within 10% of the theoretical value (ans).
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