Answer:
1. The solid is dissolved in the hot solvent (ethanol or water);
2. If the impurities are not dissolved, they are separated by filtration;
3. The solution is then cold, and the crystals of the solid are formed;
4. The solution is filtrated and the pure solid is obtained.
Explanation:
The recrystallization is a process to separate impurities from a solid. The solid with the impurities is dissolved in a hot solvent, and, is insoluble in the cold one. But the impurities must be soluble in the cold solvent, or insoluble in the hot one.
If the impurities are soluble in the cold solvent, then, the crystals of the analyte will be removed, and if they are insoluble in the hot solvent, then its crystal is removed first.
So let's assume that the solid is insoluble in hot ethanol, and soluble in hot water. Because its insoluble in hexane, the recrystallization is not possible with it. So the procedure would be:
1. The solid is dissolved in the hot solvent (ethanol or water);
2. If the impurities are not dissolved, they are separated by filtration;
3. The solution is then cold, and the crystals of the solid are formed;
4. The solution is filtrated and the pure solid is obtained.
Which element in each of the following sets would you expect to have the highest IE₂?
(a) Na, Mg, Al (b) Na, K, Fe (c) Sc, Be, Mg
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
With increase in atomic size of the atom, there will be less force of attraction between the nucleus and the valence electrons of the atom. Hence, with lesser amount of energy the valence electrons can be removed easily.
Since, Na, Mg and Al are all period 3 elements. And, when we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Hence, smaller is the size of an atom more energy is required to remove an electron.
Therefore, out of Na, Mg, the highest [tex]IE_{2}[/tex] will be that of Na. This is because when sodium will lose one electron then it forms [tex]Na^{+}[/tex] ion which is stable in nature.
Hence, in order to remove another electron from [tex]Na^{+}[/tex] will be difficult. Therefore, it will have high [tex]IE_{2}[/tex].
Similarly, Na will have highest [tex]IE_{2}[/tex] as compared to K and Fe. Also because sodium is smaller in size than K.
Since, beryllium is smallest in size as compared to Mg and Sc. Hence, Be will have the highest [tex]IE_{2}[/tex].
Final answer:
The elements expected to have the highest second ionization energy (IE₂) from the given sets are Mg for set (a), Fe for set (b), and Be for set (c), based on their electronic configurations and positions on the periodic table.
Explanation:
The student is asking about the second ionization energy (IE₂) for various sets of elements. Ionization energy is the energy required to remove an electron from an atom or ion. The second ionization energy specifically refers to the energy required to remove a second electron after one has already been removed. Generally, this energy is greater than the first ionization energy because the remaining electrons feel a greater effective nuclear charge.
For the sets given:
(a) Na, Mg, Al: Mg (Magnesium) expected to have the highest IE₂ because it will be removing an electron from a full s-orbital, which requires more energy.
(b) Na, K, Fe: Fe (Iron) is likely to have the highest IE₂ as it is a transition metal with more protons in the nucleus, resulting in a stronger attraction to the remaining electrons.
(c) Sc, Be, Mg: Be (Beryllium) should have the highest IE₂ because removing the second electron will remove a completely filled s-orbital, which is a stable configuration requiring more energy to disrupt.
(a) Write the rate law for the reaction 2A + B → C if the reaction (1) is second order in B and overall third order, –rA = ______ (2) is zero order in A and first order in B, –rA = ______ (3) is zero order in both A and B, –rA = ______ (4) is first order in A and overall zero order. –rA = ______ (b) Find and write the rate laws for the following reactions (1) H2 + Br2 → 2HBr (2) H2 + I2 → 2HI
Answer:
A. Write the rate law for this reaction
2A + B -----> C
Rate law is expressed as
-rA = kaCa^αCb^β
n = α + β
Where, ka is the rate constant;
Ca is the concentration of reactant A, Cb is the conversation of reactant B, α is the rate order for reactant A, β is the rate order for reactant B, n is the overall order.
1. β = 2, n = 3; α = 3-2 = 1
Rate law is;
-rA = ka[A][B]^2
2. α = 0, β = 1
-rA = ka[A]^0[B]
-rA = ka[B]
3. α=β=0
-rA = ka[A]^0[B]^0
-rA = ka
4. α=1, n = 0
The reaction is zero order as it is independent of the reactants.
-rA = ka
B. Rate law for the following;
a. H2+ Br2 ------>2HBr
-rA1 = ka[H2] . [Br2]
-rA2 = kb[HBr]^2
Comparing both rate, rA1= rA2
ka.[H2][Br2] = kb[HBr]^2
ka/kb = K = [HBr]^2 / [H2] [Br2]
b. H2 + I2 ------>2HI
K = [HI]^2 / [H2] [I2]
The rate laws for different reaction scenarios and specific reactions.
Explanation:(a) The rate law for the reaction 2A + B → C can be determined by examining the reaction orders for each reactant. For the given scenarios:
When the reaction is second order in B and overall third order, the rate law is -rA = k[A]2[B].When the reaction is zero order in A and first order in B, the rate law is -rA = k[B].When the reaction is zero order in both A and B, the rate law is -rA = k.When the reaction is first order in A and overall zero order, the rate law is -rA = k[A].(b) The rate laws for the given reactions are:
The rate law for the reaction H2 + Br2 → 2HBr is -rH2 = k[H2][Br2].The rate law for the reaction H2 + I2 → 2HI is -rH2 = k[H2][I2].Learn more about Rate laws here:https://brainly.com/question/35884538
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A 34.57 mL sample of an unknown phosphoric acid solution is titrated with a 0.127 M sodium hydroxide solution. The equivalence point is reached when 28.2 mL of sodium hydroxide solution is added. What is the concentration of the unknown phosphoric acid solution?
The concentration of the unknown phosphoric acid solution can be determined using the concept of stoichiometry in titrations. In this case, the concentration of the sodium hydroxide solution and the volume used at the equivalence point are known. By using the given information, the concentration of the unknown phosphoric acid solution can be calculated to be approximately 0.150 M.
Explanation:The concentration of the unknown phosphoric acid solution can be determined using the concept of stoichiometry in titrations. In this case, the concentration of the sodium hydroxide solution and the volume used at the equivalence point are known. By using the equation:
NaOH (aq) + H3PO4 (aq) → NaH2PO4 (aq) + H2O (l)
the moles of sodium hydroxide can be determined. Then, by considering the volume and concentration of the phosphoric acid solution, the concentration of the unknown phosphoric acid solution can be calculated. In this case, the concentration of the unknown phosphoric acid solution is approximately 0.150 M.
An experiment calls for you to use 100 mL of 0.25 M HNO3 solution. All you have available is a bottle of 3.4 M HNO3. How many milliliters of the 3.4 M HNO3 solution do you need to prepare the desired solution?
Answer: 7.35mL
Explanation:
C1 = 3.4M
V1 =?
C2 = 0.25M
V2 = 100mL
C1V1 = C2V2
3.4 x V1 = 0.25 x 100
V1 = (0.25 x 100) /3.4
V1 = 7.35mL
Final answer:
To prepare the desired 100 mL of 0.25 M HNO3 solution, approximately 7.35 mL of the 3.4 M HNO3 stock solution is required, with the remaining volume filled with water for dilution.
Explanation:
To prepare 100 mL of a 0.25 M HNO3 solution from a 3.4 M HNO3 stock solution, we use the dilution formula M1V1 = M2V2, where M1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, M2 is the desired concentration, and V2 is the final volume of the solution.
Here, M1 = 3.4 M, M2 = 0.25 M, and V2 = 100 mL. We need to find V1. Plugging the values into the equation:
V1 = (M2 x V2) / M1
V1 = (0.25 M x 100 mL) / 3.4 M
After calculating, we find that V1 ≈ 7.35 mL. Therefore, you need approximately 7.35 mL of the 3.4 M HNO3 stock solution to prepare the desired 100 mL of 0.25 M HNO3 solution. The rest of the volume up to 100 mL should be filled with distilled water to achieve the correct dilution.
Indicate whether the substance exists in aqueous solution (a) entirely in molecular form, (b) entirely as ions, (c) or as a mixture of molecules and ions. HF, CH3CN, NaClO4, Ba(OH)2
Answer:
In aqueous solution, HF is an acid, exist as a mixture of molecules and ions
CH3CN is none of the above, entirely in molecular form
NaCIO4 is a salt, entirely as ions
Ba(OH)2 is a base, entirely as ions
The substance exists in aqueous solution entirely in molecular form is Methyl cyanide, entirely as ions is NaClO₄ & Ba(OH)₂ and HF as a mixture of molecules and ions.
What is aqueous solution?Aqueous solutions are those solution in which water is present as a solvent and we know that nature of water is polar in water means it consist positive or negative charges.
HF is a weak acid and it shows partial dissociation only, so it is present in the mixture form of molecules and ions.Methyl cyanide is a covalent molecule not dissolve in aqueous solution and present in the molecular form.NaClO₄ is a salt and it is completely dissolve in aqueous solution and present in the ions form.Ba(OH)₂ is a strong base and it is also completely dissociates into their ions.Hence NaClO₄ & Ba(OH)₂ is completely dissociates into ions.
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n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form of the acid?
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
[tex]pK_a=8.0[/tex]
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}[/tex]
Now put all the given values in this expression, we get:
[tex]6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}[/tex]
[tex]\frac{[Deprotonated]}{[Protonated]}=0.01[/tex]
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
[tex]\frac{[Protonated]}{[Deprotonated]}=100[/tex]
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
The ratio of the protonated to the deprotonated form of the acid in the solution is 100:1.
The ratio of the protonated to the deprotonated form of the acid is given by the equation:
[tex]\[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(\text{pH} - \text{pKa})} \][/tex]
Given that the pKa of the acid is 8.0 and the pH of the solution is 6.0, we can plug these values into the equation:
[tex]\[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(6.0 - 8.0)} \] \[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2} \] \[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = \frac{1}{100} \][/tex]
Therefore, the ratio of the deprotonated form to the protonated form of the acid is 1:100. To find the ratio of the protonated to the deprotonated form, we take the reciprocal of this value:
[tex]\[ \text{Ratio of protonated to deprotonated} = \frac{[\text{HA}]}{[\text{A}^-]} = 100:1 \][/tex]
The answer is: 100:1.
Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the gene in the 5' to 3' direction.
The possible 5' to 3' sequence of bases for the sticky end of the gene cut by PpuMI is: 5'-ACGGA-3'.
The enzyme PpuMI cuts DNA at recognition sequences. In the 5' to 3' direction, "ACGGA." might be the sticky end of the gene snipped by PpuMI. This indicates that one DNA strand reads 5'-ACGGA-3' and the other 3'-TGCCT-5'.
These overhanging single-stranded ends are called "sticky ends" because they can base-pair with complementary sequences in another DNA fragment cut with the same enzyme. In genetic engineering, like cloning, DNA fragments with matching sticky ends can be joined to form recombinant DNA. PpuMI-generated sticky ends with the sequence "ACGGA" allow scientists to insert genes of interest into plasmids or other DNA molecules for study and practical applications.
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The possible sequence of bases for the sticky end of the gene in the 5' to 3' direction after being cut with PpuMI is 5'-GAAC (gene) TTGC-3'.
Explanation:The sequence of bases for the sticky end of the gene in the 5' to 3' direction after being cut with the enzyme PpuMI can be determined from the given information. The restriction enzyme PpuMI leaves a 2- to 4-nucleotide single-stranded overhang on each strand of the DNA after cutting. The sequence that is recognized by PpuMI is a palindrome, meaning it reads the same forward and backward. Therefore, the possible sequence of bases for the sticky end of the gene in the 5' to 3' direction is: 5'-GAAC (gene) TTGC-3'
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The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) Cl2(g) PCl5(g) Calculate the equilibrium concentrations of reactant and products when 0.400 moles of PCl3 and 0.400 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.
Answer:
[PCl₅] = 0.336 M
[Cl₂] = [PCl₃] = 0.064 M
Explanation:
We are given the equilibrium constant, kc , and the moles of reactants and products. Thus, our strategy here sholud be to express the quantities at equilibrium in terms of its constant by setting up our ICE table helper.
First lets start by writing the expression for the equilibrium constant:
PCl₃ (g) + Cl₂ (g) ⇄ PCl₅(g)
Kc = [PCl₅] / [Cl₂] / [PCl₃] =83.3
and setup the ICE table
We are given the moles introduced in the 1.00 L vessel, so we can calculate the molarities as M = mol/L
[Cl₂] [PCl₃] [PCl₅]
i 0.400 M 0.400 M 0
c -x -x +x
e 0.400 - x 0.400 - x x
x / (0.400 - x )² = 83.3
(0.16 - 0.8x + x²) x 83.3 = x
13.33 -66.66x + 83.3x² = x
13.33 - 67.66 x + 83.3 x² = 0
Solving the quadratic equation we have x₁ = 0.476 and x₂ = 0.336
The first solution is phisically impossible, since it will give us a negative quantity at equilibrim for the reactants.
With the second solution x = 0.336, the equilibrium concentrations are:
[PCl₅] = 0.336 M
[Cl₂] = [PCl₃] = 0.400 - 0.336 = 0.064 M
Which functional group does not contain an oxygen atom? Group of answer choices amine amide aldehyde ketone ether ester alcohol carboxylic acid
Answer: Amine does not contain an oxygen atom
Explanation:
amine is represented by -NH2
amide is represented by -CONH2
aldehyde is represented by -CHO
ketone is represented by -CO
ether is represented by -CO
ester is a represented by -COOR
alcohol is represented by -OH
carboxylic acid is represented by -COOH
You will observe that the only functional group without an oxygen atom is amine with -NH2
Amines are the functional group that does not contain an oxygen atom, distinguishing them from aldehydes, ketones, carboxylic acids, esters, ethers, alcohols, and amides. Option A is correct.
The functional group that does not contain an oxygen atom among the listed options is the amine group. Functional groups like aldehydes, ketones, carboxylic acids, esters, ethers, alcohol, and amide are characterized by the presence of oxygen in their structures.
Notably, the carbonyl group common to aldehydes, ketones, carboxylic acids, and esters features a carbon-oxygen double bond. Amines, on the other hand, are compounds containing nitrogen atoms bonded to alkyl or aryl groups, and they do not incorporate oxygen within their functional group.
Hence, A. is the correct option.
The complete question is:
Which functional group does not contain an oxygen atom? Group of answer choices
A) amine
B) amide
C) aldehyde
D) ketone
E) ether
F) ester
G) alcohol
H) carboxylic acid
The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 11.0 minutes?
Answer: The concentration of reactant after the given time is 0.0205 M
Explanation:
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]4.50\times 10^{-3}s^{-1}[/tex]
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
[tex][A_o][/tex] = initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}[/tex]
[tex][A]=0.0205M[/tex]
Hence, the concentration of reactant after the given time is 0.0205 M
Calculate the concentration of hydronium in a solution that contains 5.5x10-5 M OH- at 25C. Indentify the solution as acidic, basic, or neutral.
Answer:
pH = 9.74
The solution is basic
Explanation:
To find the pH of the solution, we need to find the pOH of the solution.
From the question, the concentration of OH^- = 5.5x10^-5 M
pOH = - Log[OH^-]
pOH = - Log 5.5x10^-5
pOH = 4.26
Recall,
pH + pOH = 14
pH = 14 — pOH
pH = 14 — 4.26
pH = 9.74
Since the pH is above 7, the solution is alkaline ie basic
You have a freshly prepared 1 M (molar) solution of glucose in water. You carefully pour out a 100 mL sample of that solution. How many glucose molecules are included in that 100 mL sample?
Answer:
[tex]6.022\times 10^{22} [/tex]glucose molecules are included in that 100 mL sample.
Explanation:
Concentration of freshly prepared glucose solution = 1 M = 1 mol/L
1 L = 1000 ml
This means that 1 mole of glucose is present in 1000 mL of water.
If we have 100 mL of solution. then number of moles of glucose will be L;
[tex]\frac{1}{1000}\times 100 mL=0.1 mole[/tex]
1 mole = [tex]N_A=6.022\times 10^{23} [/tex] molecules/atoms
Number of molecules of glucose in 0.1 mole :
= [tex]0.1 mol\times 6.022\times 10^{23} molecules=6.022\times 10^{22} moleules[/tex]
If the pH of a phosphoric acid (H3PO4) solution is adjusted to 6.50, what is the most abundant species and which is the second most abundant species? For H3PO4 Ka1 = 7.11 x 10-3, Ka2 = 6.34 x 10-8, and Ka3 = 4.22 x 10-13.
Explanation:
For the given values of [tex]K_{a}[/tex] we will have the values of [tex]pK[/tex] as follows.
As, [tex]pK_{a} = -log K_{a}[/tex]
Therefore,
[tex]pK_{a1}[/tex] = 2.15, [tex]pK_{a2}[/tex] = 7.20
[tex]pK_{a3}[/tex] = 12.38
Now, at pH 6.50
[tex]H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+}[/tex]; [tex]K_{a1}[/tex]
At pH = 2.15; [tex]H_{3}PO_{4} = H_{2}PO^{-}_{4}[/tex]
[tex]H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+}[/tex]; [tex]K_{a2}[/tex]
At pH 7.20; [tex]H_{2}PO^{-}_{4} = HPO^{2-}_{4}[/tex]
[tex]HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+}[/tex]; [tex]K_{a3}[/tex]
Hence, we can conclude that most abundant species is [tex]H_{2}PO^{-}_{4}[/tex] and the second most abundant species is [tex]HPO^{2-}_{4}[/tex].
What does ""electron density in a particular tiny volume of space"" mean?
Explanation:
Electron Density is nothing but finding of some electron in a tiny space of per unit volume. It is rather a probability of finding electron in per unit volume, it should be called electron probability density. And there is function Ψ^2 that represent probability of finding electron in some tiny volume of of the atom.
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.27 m NH4I A. Highest boiling point 2. 0.11 m FeCl3 B. Second highest boiling point 3. 0.16 m CaI2 C. Third highest boiling point 4. 0.50 m Sucrose(nonelectrolyte) D. Lowest boiling point Submit AnswerRetry Entire Group9 more group attempts remaining
Answer:
1. 0.27 m [tex]NH_4I[/tex] : Highest boiling point
2. 0.11 m
: Lowest boiling point
3. 0.16 m
: Third highest boiling point
4. 0.5 m sucrose : Second highest boiling point
Explanation:
[tex]\Delta T_b=i\times k_b\times m[/tex]
[tex]\Delta T_b[/tex] = elevation in boiling point
i = Van'T Hoff factor
[tex]k_b[/tex] = boiling point constant
m = molality
1. For 0.27 m [tex]NH_4I[/tex]
[tex]NH_4I\rightarrow NH_4^{+}+I^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]2\times 0.27=0.54m[/tex]
2. For 0.11 m [tex]FeCl_3[/tex]
[tex]FeCl_3\rightarrow Fe^{3+}+3Cl^{-}[/tex]
i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.11=0.44m[/tex]
3. For 0.16 m [tex]CaCl_2[/tex]
[tex]CaCl_2\rightarrow Ca^{2+}+2Cl^{-}[/tex]
, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be [tex]3\times 0.16=0.48m[/tex]
4. For 0.5 m sucrose
, i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be 0.5 m
Thus as concentration of solute follows the order : [tex]NH_4I[/tex] > sucrose > [tex]CaCl_2[/tex] > [tex]FeCl_3[/tex], the boiling point will also follow the same order.
The exponents in a rate law can be found:
Select the correct answer below:
O from the stoichiometric coefficients of the reaction
O from the molar masses of the compounds
O by experiment only
O none of the above
Answer:
by experiment only
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
Thus, the exponents in the rate law, the order is determined by experiment only.
Exponents in a rate law can only be determined by experiment, not from the stoichiometric coefficients of the reaction or the molar masses of the compounds.
Explanation:The exponents in a rate law cannot be determined from the stoichiometric coefficients of the reaction or the molar masses of the compounds. The correct answer is that they can be determined by experiment only. This is because the rate law and its exponents are indicative of the reaction mechanism, or the step-by-step sequence of elementary reactions by which overall chemical change occurs. These can't usually be predicted just from the overall reaction equation (which provides the stoichiometric coefficients) and require experimental data for determination.
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Which element would you expect to be more metallic?
(a) S or Cl (b) In or Al (c) As or Br
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.
This means that chlorine (S) is less metallic than chlorine (Cl).
Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.
Therefore, aluminium (Al) is more metallic than indium (In).
Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.
Final answer:
The most metallic element can be determined by looking at their positions in the periodic table.
Explanation:
The most metallic element can be determined by looking at their positions in the periodic table. Metallic character generally increases going down a group and decreases going across a period. Using this information, we can predict that:
(a) S is more metallic than Cl because S is below Cl in the same group and is further down the periodic table.
(b) In is more metallic than Al because In is below Al in the same group and is further down the periodic table.
(c) Br is more metallic than As because Br is to the left of As in the same period and closer to the metals in the periodic table.
How is the molar heat of sublimation related to the molar heats of vaporization and fusion? On what law are these relationships based?
Answer: Hsub=Hfus+Hvap
Explanation:
The molar heat of vaporization measured in kilojoules per mole, or kJ/mol is the energy needed to make vapor one mole of a liquid. .
The molar heat of sublimation measured in kilojoules per mole, or kJ/mol is the energy needed to sublime one mole of a solid,
the molar heat of fusion measured in kilojoules per mole, or kJ/mol is the energy needed to melt one mole of a solid.
Hess law helps to explain the relationship in physical chemistry stating that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps.
In this context Hess’s law helps to see the several steps involved as the heat of sublimation energy is equal to the sum of vaporization energy and fusion energy.
The molar heat of sublimation is the sum of the molar heats of vaporization and fusion, based on Hess's Law. This law enables understanding that sublimation is similar to a sequential process of melting followed by vaporization, reflecting the energy required to overcome intermolecular forces.
The molar heat of sublimation is directly related to the molar heats of vaporization and fusion. The relationship between these heats is guided by the Hess's Law, which asserts that the total enthalpy change during a chemical reaction is the same no matter how many steps the reaction is carried out in.
Thus, for a given substance, the molar heat of sublimation can be approximated by the sum of its molar heat of fusion (solid to liquid transition) and molar heat of vaporization (liquid to gas transition). This is because when fusion is followed by vaporization at the temperature and pressure of the triple point, the net change corresponds to sublimation.
The molar enthalpies related to phase changes are all positive for endothermic processes (where heat is absorbed). Conversely, negative enthalpy changes are associated with the reverse, exothermic processes. The differences in these heats reflect the extent to which intermolecular forces must be overcome going from one phase to another.