Heat energy xxxxxxxxxxxxxx
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
N2(g) + 3H2(g) =2NH3(g)
Number of moles of reactants > product...
Therfore if pressure is increased.
Equilibrium is disturbed according to LCP....
Equilibrium shift to the right (product)
......
If you want to understant the fundamental concept:
Take for example 2A + 3B = 4C
Reactant: 5 moles (5 volumes)
Product: 4 moles (4 volumes)
When pressure of a gas is increased, volume decreases!
(Vice-versa)
If pressure is increased, volume decreases. Hence number of collisions INCREASES(constrain). Equilibrium shifts in such a direction so as to decrease the number of collision accordinf to LCP...
This happens when number of paeticles decreases as equilibrium shift forward because the forward reaction is accompanied by a decrease in number of particles (5 to 4)