Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
A word equation is a chemical reaction described using words.
A common example is the act of photosynthesis - the process plants use to make glucose (sugar) to use as 'food'.
Plants convert water and carbon dioxide into oxygen and glucose.
A word equation to express this is:
Water + Carbon Dioxide → Glucose + Oxygen
The other type of equation is a symbol equation - this uses the symbols of the elements instead of the common names:
H₂O + CO₂ → C₆H₁₂O₆ + O₂
There is also a balanced version:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
<em>If you want information on the balanced symbol equations, feel free to PM me.</em>
Answer:
Coefficient = 1.58
Exponent = - 5
Explanation:
pH = 2.95
Molar concentration = 0.0796M
Ka = [H+]^2 / [HA]
Ka = [H+]^2 / 0.0796
Therefore ;
[H+] = 10^-2.95
[H+] = 0.0011220 = 1.122 × 10^-3
Ka = [H+] / molar concentration
Ka = [1.122 × 10^-3]^2 / 0.0796
Ka = (1.258884 × 10^-6) / 0.0796
Ka = 15.815 × 10^-6
Ka = 1.58 × 10^-5
Coefficient = 1.58
Exponent = - 5
Compare HCl, NaOH, and NaCl: HCl is a stronger acid than water. NaCl is a weaker base than NaOH. Strong acids react with strong bases to form weaker acids and bases. ... Compare NaOH, NH3, and H2O, and NH4Cl: NaOH is a stronger base than NH3.
