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
They define acids as proton donors, and bases as proton acceptors
If you were to have:
HNO3 + H2O -> H3O+. + NO3-
You can see that the nitric acid (HNO3) gave a hydrogen ion which has 1 proton, 0 neutrons and 0 electrons to the water so we just say that it gave a proton.
Now let's see a base
NH3 + H2O -> NH4+ + OH-
Now, you can see that the ammonia (NH3) gained a hydrogen ion (proton) from the water to become ammonium(NH4). which means it accepted a proton
That's basically it. Feel free to ask if you have any further questions
13
Al
Aluminum
Atomic mass 26.982
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