Answer:
Keq = 0.053
7.3 kJ/mol
Explanation:
Let's consider the following isomerization reaction.
glucose 6‑phosphate ⇄ glucose 1 - phosphate
The concentrations at equilibrium are:
[G6P] = 0.19 M
[G1P] = 0.01 M
The concentration equilibrium constant (Keq) is:
Keq = [G1P] / [G6P]
Keq = 0.01 / 0.19
Keq = 0.053
We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.
ΔG° = -R × T × lnKeq
ΔG° = -8.314 J/mol.K × 298 K × ln0.053
ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol
Hi!
The best possible name for this molecule would be: 4 - ethyl - 1 - heptene or 4 - ethylhept - 1 - ene
We always name the molecule with respect to the longest chain (in a branched molecule), taking both atoms of carbon participating in the double bond into account too. In our case, this gives us a 7 Carbon chain - hept
ene - is the suffix that is indicative of the molecule being an alekene.
<em>So we know it is a branched heptene molecule</em>
We add the number where the double bond occurs either before ene, or before heptene as a rule.
<em>Note: We always start with the end of the chain from where the double bond is the closest, and number the carbons accordingly.</em>
The title and position of the branch always comes at the start. In our case the branch is a two carbon chain, and an alkane, so it would be an ethyl branch. This branch occurs at carbon number 4
Hence, the correct names would be:
<em>4 - ethyl - 1 - heptene</em> or <em>4 - ethylhept - 1 - ene</em>
Hope this helps!
