Heat of Chemical Reaction Problems It is known that the temperature of the solution decreases when b

1mol anhydrous copper sulfate is made into a solution when the heat is Q1KJ, and 1mol of bile alum decomposition will absorb the heat of H=+Q2KJ, 1mol of bile alum dissolved in water first of all to break the chemical bond between CuSO4 and 5H2O, absorb heat Q2, CuSO4 ionization and release Q1 heat, so 1mol of bile alum dissolved in water will absorb Q2-Q1 heat.

It is known that the temperature of the solution decreases when the bile alum is dissolved in water (this refers to the condition of not heating), indicating that Q2 is greater than Q1 and the process is endothermic as a whole. If it does not absorb heat, the internal energy decreases after dissolution and the temperature drops (the actual process may be endothermic, but the heat transfer is slower, so the temperature drop will still be observed).

Q2 is approximately equal to the energy required to break the chemical bond between Cuso4 and 5H2O, but it is only an approximation, and the reason is very complicated, and middle school students do not need to figure it out. Q2 does not include the amount of copper sulphate that is dissolved in water and exothermic.

If you have any questions, please feel free to ask.

According to the description of your question, the dissolution of bile alum in water can be regarded as two processes, the first process is: cuso4 5h2o= =cuso4 (s)+5h2o; △h=+q2kj/mol

This process releases heat, and the heat absorbed by the decomposition of one mole of CuSO4 5H2O is Q2KJ;

The second process is: the process of making anhydrous copper sulfate into a solution, this process is to release heat, 1mol of anhydrous copper sulfate to make a solution when the heat is released is Q1kj, but the whole process should reduce the temperature of the solution, indicating that the absorbed heat is greater than the heat released, so Q2>Q1

Compare the size of the calories (in question.)"If 1mol of anhydrous copper sulfate is dissolved into a solution, it is exothermic q2kj")

When anhydrous copper sulfate is dissolved into a solution, first exothermic Q1 kj (becomes pentahydrate) and redissolves pentahydrate.

The total put q2, q2 is the value of q1 put minus a certain endothermic value.

So Q1>Q2, choose A

Satisfied? I want to give points).

You're a high school student ......I'm a junior high school student and can't read ......=

The dissolution cooling indicates that the heat release of hydration is less than the biliang heat of diffusion absorption, and Q1 is the difference between the two.

And the fraction of gallium (pure copper sulfate pentahydrate) is to overcome the hydration energy q2, so.

q2>q1

q2 is equal to q1, and the heat of reaction in the stem is -q2, and they are equal in value.

When the temperature of the solution decreases when the cholelum is dissolved in water, it can be seen that it is an endothermic process, and when H 0, 1mol of bile alum decomposes to 1mol CuSO4(S), CuSO4?5h2o（s）△

cuso4（s）+5h2o（l）△h=+q1 mol?When L-11molCuSO4(S) is dissolved, the exothermic Q2 kJ can be obtained as shown in the figure below, according to Geis's law, then there is H=Q1+(-Q2) 0, then Q1 Q2, so choose A

2. After absorbing the heat of osmotic reaction (chemical reaction and such as heat absorption), there is a selection of D, heating the gallum into white powder, potassium chlorate decomposition and slag to take oxygen (most of the decomposition reactions are endothermic, and these two reactions must be continuously heated).

Endothermic but not reaction, Exothermic.

The chemical equation for the solution of bile alum heating is:

Heating lead noisy == cuso4 + 5h2o

This is the water reaction of heating loss.

Bile imitation hail alum. The chemical formula for the spare sail is CuSO4·5H2O

After it dissolves in water, it is actually a CuSO4 solution mold roll, so the solute is: CuSO4

== Heating == Cuso4 + 5H2O

This is the heat loss reaction.

==Heating==

cuso45h2o

This is the heat loss reaction.

Decomposition Phenomenon: The blue crystals gradually turn into white powder, and water droplets are formed at the mouth of the test tube.

Water and copper sulphate are generated.

Copper oxide can be formed by overheating.

Chemical equation for thermal decomposition of bile alum:

Heating ==cuso4

5H2O heating water loss reaction decomposition phenomenon: blue crystals gradually turn into white powder, and water droplets are generated at the mouth of the test tube.