Determine the enthalpy of solution for a solid with δh values as described in each scenario.
2 answers:
Box 1, the value of ΔH solution is positive (endothermic)
Box 2 ΔH value = 0 and ,
Box 3 ΔH the value of ΔH solution is negative (exothermic)
<h3>Further explanation </h3>Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The enthalpy of solution formation depends on how much heat is released or absorbed during the dissolution process. So it can be negative (exothermic) or positive (endothermic)
When the solid is dissolved into the solution, 3 stages of the process will occur:
- 1. The enthalpy needed to break the bonds of molecules dissolved solid compounds
This enthalpy is endothermic, can be symbolized as ΔH1
- 2. The enthalpy needed to break the bonds of solvent molecules, this enthalpy is also endothermic, ΔH2
- 3. The enthalpy needed to combine the bond between solute and solvent, this enthalpy is exothermic, ΔH3
If the enthalpy of stage 1 + 2 which is a bond-breaking enthalpy is greater than the enthalpy of stage 3 which is the bond-forming energy, the enthalpy of the solution will be endothermic (positive value,> 0), but vice versa if the enthalpy of stage 1 + 2 is smaller than the stage 3, the enthalpy of the solution will be exothermic (negative, <0)
And if stage 1 + 2, the enthalpy value is the same as stage 3, the enthalpy of the solution is 0, and it is said that this solution is ideal
From the above problem, we complete the picture that is asked according to the scenario
If we look at box 1, the value of ΔH of the solution will be positive because the enthalpy of the bond breaker is 3x greater than the enthalpy of the bond-forming, so the solution is endothermic
For box 2, the solution ΔH is 0 because the enthalpy values 1 + 2 and 3 are the same
As for box 3, ΔH solution will be negative (exothermic) because enthalpy 1 + 2 is less than 1/3 of enthalpy stage 3
<h3>Learn more </h3>
an exothermic reaction
as endothermic or exothermic
an exothermic dissolving process
Keywords: exothermic, endothermic, enthalpy
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Answer:
1. Stąd empiryczny wzór substancji to N₂O₅
2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.
Explanation:
1. Mamy tutaj;
Masa molowa tlenku azotu = 108u
Masa azotu = 14,0067u
Masa tlenu = 15,999 u
74,07% masy tlenku azotu to tlen
Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u
Masa obecnego azotu = 108 - 79,9956 = 28,0044u
Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5
Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2
Stąd empiryczny wzór substancji to N₂O₅.
2. Kiedy sód reaguje z chlorem, mamy;
2Na (s) + Cl₂ (g) → 2NaCl (s)
Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl
W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl
Masa Na obecnego w reakcji = 45 g
Masa molowa sodu = 22,989769u
Liczbę moli sodu w 45 g sodu podano w następujący sposób;
Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl
Masa molowa NaCl = 58,44 g / mol
Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl
Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g
W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.