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Green checks and red X's are not displayed for parts b, c and d of this question. A pair of students found the temperature of 100. g of water to be 27.80°C. They then dissolved 8.52 g of LiBr in the water. When the salt had dissolved, the temperature of the water was 39.25°C.
a) Calculate ΔT for the water.
b) The dissolution wasexothermic.
c) The waterabsorbed energy from the dissolution process.
d) Based on this observation alone, the entropychange cannot be determined.
e) Give the reaction for the dissolution of the salt in water. (Use the lowest possible coefficients. Include states­of­matter under the given conditions in your answer.)

Respuesta :

Explanation:

a) Initial temperature of the water = [tex]T_1=27.80^oC[/tex]

Final temperature of the water = [tex]T_2=39.25 ^oC[/tex]

Change in temperature of the water = ΔT

ΔT = [tex]T_2-T_1[/tex] =  39.25°C - 27.80°C = 11.45°C

b) During an exothermic reaction energy is released into surrounding which results in an increase in temperature of the surroundings.

So, when students mixed LiBr in water the final temperature of the water was more than that of the initial temperature which means that dissolution was exothermic.

c) During the dissolution process the temperature of the water rises up which means that water must have absorbed heat from the dissolution reaction.

d) On dissolution of solid LiBr in water breaks the solid LiBr into ions which are independent to move freely in solution due to which randomness of the system increases and hence, Entropy system has increased as the number of particles on dissolution has increased.

e) The reaction for the dissolution of the salt in water:

[tex]LiBr(aq)\rightarrow LI^+(aq)+Br^-(aq)[/tex]

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