Answer:
Explanation:
1 mol of anything is 6.02 * 10^23 atoms (in this case)
x mol of 3.01 * 10^22
Set up the proportion
1/x = 6.02*10^23 / 3.01 * 10^22 Cross multiply
x*6.02 * 10^23 = 1 * 3.01 * 10^22 Divide by 6.02*10^23
x = 3.01 * 10^22 / 6.02*10^23
x = 1/(2 * 10)
x = 1/20 mol
x = 0.05 mol
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