∆Grxn 0 = ∑[∆Gf(S) 0 + ∆Gf(O2) 0  ]  - 2∆GSO3 0 =  [2*0 +0] –[2*-370.4]


Now ΔGf0of any element in its stable form is zero.
For S and O₂ it is zero and plug in the value of SO₃ to get the answer.

It is from reactant to the top of the hill.Can you post the complete question?

Hi mam,

I wanted to ask why can't option A be the one with the highest activation energy

hi mam,

I am confused on how to do this problem!

Draw the Lewis structure of ethyne, C₂H₂, first:

H–C≡C–H

Each carbon must make four bonds to satisfy its tetravalency. In ethyne, each carbon is already bonded to one hydrogen, so each carbon still needs three more bonds. Therefore, the two carbon atoms must be connected by a triple bond.

Now remember:

A triple bond consists of one sigma bond and two pi bonds.

So, in one molecule of ethyne, there are 2 pi bonds.

ΔEN cannot be the correct answer here.
The electronegativity difference in S–Cl is actually less than in Be–Cl. So, based only on ΔEN, BeCl₂ should have the more polar bonds. That part is true.
However, bond polarity and molecular polarity are not the same thing.
In BeCl₂, each Be–Cl bond is polar, but the molecule is linear:
Cl — Be — Cl
Because the two identical bond dipoles are equal and point in opposite directions, they cancel out completely. So the net dipole moment is zero.
In SCl₂, sulfur has two lone pairs as well as two bonding pairs. Its electron geometry is tetrahedral, but its molecular shape is bent (V-shaped), not linear with two lone pairs on S
Because the molecule is bent, the two S–Cl bond dipoles do not cancel. So SCl₂ has a nonzero dipole moment.
ΔEN tells you whether an individual bond is polar, but it does not by itself tell you whether the whole molecule is polar.
To determine molecular polarity, you must draw the Lewis structure, find the shape of the molecule, and check whether the bond dipoles cancel due to symmetry.

Hi Mam,

For this question I was wondering why can't electronegativity differences also be a right answer when considering dipole movement?

Hi Mam,

For this question doesn't ethyne have only single bonds so it doesn't have any PI bonds?

Rate determining step is slo
so rate = k[H2]^2
However I is the intermediate and rate law is always in terms of reactant of the overall reaction.
S we need to replace intermediate by the reactant from the fast equilbrium step which is I₂
 rate = k[H2][I₂]

Key Concept (AP Chemistry level)
Intermediates (like
𝐼
I) must be eliminated.
Use equilibrium expression of fast step to substitute.
Final rate law must contain only reactants from overall reaction.

I am also confused about how to draw a energy diagram of a reverse uncatalyzed rxn?