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Question 1
According to “Hess’s Law”, energy cannot be gained or lost; however, it can only be changed from one form to another form. This law is known as the “Law of conservation of Energy” where he claimed that Energy can only Chemical Energeticsbe conserved and neither be made nor be destroyed.
The importance of measuring enthalpy is to determine whether the reaction has absorbed the heat or released the heat. Therefore, if the reaction absorbed the heat it is known as an endothermic reaction or “positive change in enthalpy”. In addition, if the heat is released at the end of the reaction, it is known as an exothermic reaction or “negative change in enthalpy”. Other than that, it is also used to measure the amount of heat absorbed or released and that is measured in calorimetry. In addition, it is used to evaluate “Joule-Thompson expansion” or “throttling process”. Hence, due to all these reasons, it is important to understand the chemical processes (Li et al. 2019).
H2 + has a delta H value of -288, and in the first equation there are 6 H2O. Therefore, 6 x>
Therefore, here +>
Since there are 4 NH3,>
Question 5
B2H6(g) + 3O2(g) -> B2O3(s) + 3H2O(g)
+31.4 + 0 -> -1279 +>
C6H6(g) + 7.5O2(g) -> 6CO2(g) + 3H2O(g)
+83.9 + 0 ->6*(-393) +>
Question 6
C2H5OH + 3O2 -> 2CO2 + 3H2O + 1368 kJmol-1
2 moles of 2CO2, therefore,>
3 moles of so H2O, therefore,>
Hence,>
Question 7
CH4(g) + Br2(g) -> CH3Br(g) + HBr(g)
The ΔH for this reaction is -71.7Kj/mol – (-74.6 kj/mol
CH4(g) + F2(g) -> CH3F(g) + HF(g)
[4(413) + 1(155) - 3(413) + 1(485) + 1(567)]
[(1652 + 155) - (1239 + 485 + 567)]
[1807 - 2291]
484 KJ/mol
Enthalpy of formation of HF
396.7 j/(mol.K)- 388.8 J/(mol K) + 7.9J/ (mol K)
enthalpy of formation of propane
3C+3H2?C3H6 kJ mol−1.6
C+O2?CO2 kJ mol−1.
H2+0.5 O2?H20 kJ mol−1.
2 +3ΔHf(H20)−ΔHf(C3H6)]
C2H6 + Br2 -> 2(CBrH2)
ΔrH°(298.15 ± 0.30 kcal/mol
C(s) + /> 2H2(g) + /> CH4(g) +>
C(s) + O2(g)?CO2(g) (1)
2H2(g) + O2(g)?2H2O(l) (2)
CH4(g) + 2O2(g)?CO2(g) + 2H2O(l)>
CO2(g) + 2H2O(l)?CH4(g) + 2O2(g) (3)
Adding up above (1), (2), & (3) equation, it produces-
C(s) + 2H2(g) ? CH4(g)
Formation of methane is relevant from the previous equation; therefore, enthalpy for formation of methane is-
- 571.8) + 890.3
+ 890.3
kJ
2C(s) + 2O2(g) →2CO2(g) →2CO2(g), (1)
3H2(g) + 3/2 O2(g)→ 3H2O(l), (2)
Now add the equation 1 and 2, give this following equation-
2C(s) + 3H2(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l) (3)
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l), (4)
Now, subtract equation 4 from equation 3 give the following equation-
2C(s) + 3H2(g) → C2H6(g)
Therefore, enthalpy for formation of ethane is-
– (-1560.0)
kJ
Equation for the heat combustion of ethane is-
C2H6(g) + 3½O2(g) → 2CO2(g) + 3H2O(l) + heat
ΔHf[products] - Σ ΔHf[reactants]
Þ X ΔHf(CO2) + 3 X ΔHf(H2O) - ΔHf(C2H6) – 3½ X ΔHf(O2)
X (-393.5 kJ mol-1) + 3 X (-285.8 kJ mol-1) – (- 84.7 kJ mol-1) – 2 X (0 kJ mol-1)
1559.7 kJ mol-1
Question 8
N2(g) + 3H2 (g) -> 2NH3> + 3*436) – * 46
+ 3*436) + 2 *>>
Enthalpy diagram from N-H bond is seen below-
According to this equation and enthalpy calculation, formation of ammonia is an endothermic equation because enthalpy change in the product is negating in comparison with reactants.
Question 9
(+715+242-135.5)>
The average bond dissociation enthalpy of C-Cl bond is 273.833 kJmol-1
Question 10
Enthalpy of formation of 1-iodobutane is -52.0 kJmol-1
Bond dissociation enthalpy of I-I is +214 kJmol-1
4C(s) + 4.5H2 + 0.5I2 --> C4H9I ..... kJ
Bond dissociation enthalpy of C-C is +347 kJmol-1
Bond dissociation enthalpy of H-H is +435 kJmol-1
Bond dissociation enthalpy of the C-I bond is-
>reactant bond−∑Hproduct bond
4.5*435 + 0.5*214) – (2*347+9*413+x)
– (4411+x)
>>
Bond dissociation enthalpy of C-I bond is -1573.5 kJmol-1
Question 11
Bond dissociation enthalpy of the C-O bond is 108.3 kcal/mol
Question 12
When the zinc reacts with copper sulphate (CuSO4) solution, the gets replaced by the zinc sue to its more reactivity than zinc and at the end of the reaction, zinc sulphte (ZnSO4) as well as Copper forms and that turn the blue coloured solution into a colourless solution.
In this regard, change in the enthalpy is>
Question 13
C2H5OH(l) +3O2 (g)?2CO2(g)+3H2O(l)
The combustion of ethanol with the oxygen gives two molecule of carbon-di-oxide and three molecules of water.
Question 14
“Lattice dissociation enthalpy” is referred to as the change in the enthalpy that is needed to convert 1 mole of any solid crystal such as sodium into scattered gaseous ion. It is always positively charged and in case of NaCl it is +787 kJ mol-1. NaCl which is in solid state when dissociates is converted into Na+ (g) and Cl- (g).
“Enthalpy of ion hydration” is defined as energy released following the dilution of one mole of ion, which is in the gaseous state. This is an exothermic reaction, where energy is released due to new bond formation that occurs in between water and ion. In case of NaCl, Na+ (g) and Cl- (g) form bond with water molecule. Here Cl- forms bond with hydrogen ion and Na+ forms bond with OH- and in this way NaCl dissolves in water (Yang et al. 2018).
Question 15
In order to dissolve any ionic compound in water, it becomes very crucial to supply energy that helps in breaking ion lattice. However, energy is released when new bonds are formed between a compound and water molecule. In addition, lattice enthalpy is the heat that is involved when a lattice is made from any gaseous ions. Lattice energy is required in order to convert one mole ionic solid into its gaseous form. This process is highly endothermic as it absorbs energy.
[ΔLEHØ] Na+ (g) +Cl- (g) NaCl (S)
The hydration enthalpy is defined as the change of enthalpy when gaseous molecules dissolve in sufficient moles of water in order to change into a solution. In case of NaCl, 4 to 8 water molecules directly co-ordinate around the cation (Na+) and anion (Cl-). This makes the ions become neutral and hydration enthalpy helps them to dissolve in water. Caesium is much larger in size than sodium and this is the reason it has a less densecharge density and sodium being smaller in size, has a higher charge density (Amend and LaRowe, 2019). This determines the number of water molecule in the hydration cell and this is the reason why sodium ion has a much larger unit than a cesium ion.
Δhyd HØ∝, Where Z2 is the charge of the ion and r is the radius of the ion. Enthalpy change of hydration become more negative or exothermic as the radius decreases and charge increases because of the charge density of the ion increases, more ion dipole ions are formed between the ion and the solvent, which is the water in this case. Hence, smaller the ionic radius, the more is the exothermic the enthalpy change of hydration (March-Rico et al. 2020). This is the reason, enthalpy changes relate to the solubility difference between sodium chloride and caesium chloride.
Li, C., Yin, J., Odbadrakh, K., Sales, B.C., Zinkle, S.J., Stocks, G.M. and Wirth, B.D., 2019. First principle study of magnetism and vacancy energetics in a near equimolar NiFeMnCr high entropy alloy. Journal of Applied Physics, 125(15), p.155103.
March-Rico, J.F., Huang, G. and Wirth, B.D., 2020. The effect of local chemical environment on the energetics of stacking faults and vacancy platelets in α-zirconium. Journal of Nuclear Materials, 540, p.152339.
Yang, J., Yuan, Z., Liu, X., Braun, S., Li, Y., Tang, J., Gao, F., Duan, C., Fahlman, M. and Bao, Q., 2018. Oxygen-and water-induced energetics degradation in organometal halide perovskites. ACS applied materials & in
