Get free samples written by our Top-Notch subject experts for taking the No.1 Assignment Help company in the UK.
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).
2C2H6 (g) + 7O2 (g) 4CO2 (g)+ 6H2O (l) + Δ(-83.6)
C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l)+ Δ (-285.5)
2H2 (g) + O2 (g) 2H2O
2C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) + Δ (-83.6)
2C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) + Δ (-83.6)
2H2 (g) + O2 (g) 2H2O
C(s) + Cl (g) + H2 (g)
CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g)
For this equation, kJmol-1, kJmol-1, kJmol-1
H for this reaction is-
>f(products) - ∑ΔH°f(reactants) ((-74.8)+0)> kJmol-1
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(l),>-1
H2(g) + 1/2O2(g) -> H2O(l), kJmol-1
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,>
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) +>
C2H5OH + 3O2 -> 2CO2 + 3H2O + 1368 kJmol-1
2 moles of 2CO2, therefore,>
3 moles of so H2O, therefore,>
The ΔH for this reaction is -71.7Kj/mol – (-74.6 kj/mol
[4(413) + 1(155) - 3(413) + 1(485) + 1(567)]
[(1652 + 155) - (1239 + 485 + 567)]
[1807 - 2291]
396.7 j/(mol.K)- 388.8 J/(mol K) + 7.9J/ (mol K)
3C+3H2?C3H6 kJ mol−1.6
C+O2?CO2 kJ mol−1.
H2+0.5 O2?H20 kJ mol−1.
ΔrH°(298.15 ± 0.30 kcal/mol
C(s) + /> 2H2(g) + /> CH4(g) +>
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
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-
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
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.
The average bond dissociation enthalpy of C-Cl bond is 273.833 kJmol-1
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)
Bond dissociation enthalpy of C-I bond is -1573.5 kJmol-1
Bond dissociation enthalpy of the C-O bond is 108.3 kcal/mol
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>
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.
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.
visit us for more information about Chemical Engineering Assignment Help.
Amend, J.P. and LaRowe, D.E., 2019. Minireview: demystifying microbial reaction energetics. Environmental microbiology, 21(10), pp.3539-3547.
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
Critical Evaluation Of Micro And Small-Medium...Read more
Summative Individual IT-Based Coursework Sample Profitability and risk Get...Read more
Cancer Immunology Assignment Sample Introduction - Cancer Immunology...Read more
Cyber Weapons Cause Unacceptable Blowback Introduction - Cyber Weapons...Read more
Talent Management Assignment Sample Task -1: What are the key challenges and...Read more
Systematic Literature Review On Exploring The Boundary Spanning Role Of Middle...Read more
Get your doubts & queries resolved anytime, anywhere.
Receive your order within the given deadline.
Get original assignments written from scratch.
Highly-qualified writers with unmatched writing skills.
Ph.D. Writers For Best Assistance
No AI Generated Content
offer valid for limited time only*