1/ A spirit burner containing propan-1-ol is used to heat up a beaker of water
(400cm3)from 20°C to 70°C.
(Given: Specific heat capacity of water (c)=4.2Jg-1K-1;density of water=1gcm^ -3
;standard enthalpy change of formation of [propan-1-ol(l)]= -2010kJmol^ -1)
a/ Calculate the theoretical mass of propan-1-ol needed for the heating process.
b/ Would you expect a larger or smaller actual mass of propan-1-ol needed?
c/ Explain your answer in part(b).
2/A65.00cm3 sample of 0.5M sulphuric acid at 25.0°Cwas mixed with 50.00cm 3 of 0.90M potassium hydroxide solution at 25°C in a simple calorimeter.The
highest temperature recorded after mixing was 30.3°C
(Assume that the specific heat capacity and the density of the resultant solution are 4.2Jg^ -1K^ -1 and 1.0gcm-3 respectively,and there is no heat lost to the
surroundings.)
a/ Write an equation for the reaction.
b/ Define 'standard enthalpy change of neutralization'.
c/ Write an equation that can represent standard enthalpy change of
neutralization between sulphuric acid and potassium hydroxide.
d/ Determine standard enthalpy change of neutralization between sulphuric acid and potassium hydroxide.
3/ When 80.0g of sodium hydroxide(NaOH) was added to 850cm3 of water at
23.0°C, the temperature of the water rose to28.5°C after the NaOH had
dissolved.
(Assume that the specific heat capacity and the density of the resultant solution are 4.2Jg^ -1K^ -1 and 1.0gcm^ -3 respectively, and there is no heat lost to the
surroundings.)
a/ Is the dissolving process of NaOH endothermic or exothermic?
b/ How much energy is absorbed by the water?
c/ Hence, calculate the standard enthalpy change of solution of NaOH.
Chem..Standard enthalpy change
2010-10-28 3:23 am
回答 (1)
2010-10-29 6:15 am
✔ 最佳答案
1a. heat required to heat up the water
= mc(delta)T
= (400x1) x 4.2 x (70-20) = 84000 J
but then, the question stuck here. you only gave enthalpy change of formation of propanol, which is insufficient for calculation.
you'll need enthalpy change of combustion or something like that.
bc. larger mass, as in the burning, heat is lost to surrounding.
or you can state the law of thermodynamics: energy conversion cannot be 100% efficient.
2a. shouldn't be difficult.
b. refer to your notes. you should know that.
c. KOH + (1/2)H2SO4 -----> (1/2)K2SO4 + H2O
the reason to write like this is related to definition of "enthalpy change of neutralization". once you look up for it, you'll understand.
d. no. of mole of water produced = 50/1000 x 0.9 = 0.045 mole
(sulphuric acid is in excess.)
heat produced = mc.deltaT
= (65+50)x1 x 4.2 x (30.3-25) = 2559.9 J
enthalpy change = -2559.9 / 0.045 = -5.69 kJ/mol
3a. exothermic.
b. E = mc.detlaT = 850x1 x 4.2 x (28.5-23) = 19635 J = 19.635 kJ
c. no. of mole of NaOH = 80.0 / (23+16+1) = 2.0 mole
standard enthalpy change of solution = -19.6/2 = -9.8 kJ.
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