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
Answer: A reversible reaction is a reaction that takes place in back and front directions. If the reaction were to reach equilibrium, the rate of forward direction would be equal to that of the reverse reaction.
Explanation:
Reversible reactions :
These are the reaction in which reactants reacts to give product and products reacts to give reactants as a product in return.

- In above equation, 'A' and 'B' are reacting together to give 'C' and 'D', as products and vice-versa.
- When the above reaction reaches equilibrium the rate of forward and backward reaction becomes equal.
2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.
The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance. The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene. so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene
Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.
To learn more about t-butoxide here
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