The reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
<h3>What is metal displacement reaction? </h3>
Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.
As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.
Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.
Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
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Evaporating? But that’s with a boiling liquid
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
<span>To solve this we assume that the gas inside the balloon is an ideal </span>gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant volume pressure and number of moles of the gas
the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:
T1/P1 = T2/P2
P2 = T2 x P1 / T1
P2 = 25 x 29.4 / 75
P2 = 9.8 kPa
