If Qc>Kc then there will be a net backward reaction but still forward reaction will take place. My problem is if Qc>Kc forward reaction will become non spontaneous right. So how can that happen, does it like the gibbs free energy which released during backward reaction absorbs by products for forward reaction?
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2$\begingroup$ Please define Qc and Kc $\endgroup$– Ian BushCommented Jun 8, 2023 at 7:56
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1$\begingroup$ Molecules do not know Qc/Kc ratio. nothing tells them "Now you can react." nor "Now you cannot." // Trolleybuses do not spontaneously climb uphill. Yet 2 or 3 ones going downhill can create by recuperation enough energy for one going uphill. $\endgroup$– PoutnikCommented Jun 8, 2023 at 9:10
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1 Answer
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A Graph
I haven't seen this graph anywhere, so I thought it would be interesting to draw it. Consider a simple reversible reaction:
$$ \ce{A <=>[k_\text{f}][k_\text{b}] B} $$
| Time | $\ce{[A]}$ | $\ce{[B]}$ |
|---|---|---|
| $0$ | $1$ | $0$ |
| $t$ | $1-\xi$ | $\xi$ |
| $t_\text{eq}$ | $\dfrac{k_\text{b}}{k_\text{f}+k_\text{b}}$ | $\dfrac{k_\text{f}}{k_\text{f}+k_\text{b}}$ |
Reaction quotient at time $t$ is defined as
$$ Q_\text{c} = \dfrac{\ce{[B]}}{\ce{[A]}}=\dfrac{\xi}{1-\xi} \tag{1} $$
Forward and backward rates at time $t$ are given by:
$$ R_\text{f} = k_\text{f}\ce{[A]} = k_\text{f}(1-\xi); R_\text{b} = k_\text{b}\xi \tag{2} $$
From Equations (1) and (2):
$$ R_\text{f} = \dfrac{k_\text{f}}{Q_\text{c}+1}; R_\text{b} = \dfrac{k_\text{b}Q_\text{c}}{Q_\text{c}+1} $$
You can play around with this interactive plot.
Some More Clarification
There is a reason we draw the double, forward-backward arrow in $\ce{A <=>[k_\text{f}][k_\text{b}] B}$: it shows that the reaction proceeds in both directions simultaneously.
- When $Q_\text{c}<K_\text{c}$, forward reaction rate is greater than backward reaction rate:
$$ \ce{A <=>>[k_\text{f}][k_\text{b}] B} $$
- When $Q_\text{c}>K_\text{c}$, backward reaction rate is greater than forward reaction rate:
$$ \ce{A <<=>[k_\text{f}][k_\text{b}] B} $$
Equilibrium Conditions
Equilibrium is achieved when forward and backward rates are equal:
$$ R_\text{f} = R_\text{b}\implies k_\text{f}\ce{[A]_\text{eq}} = k_\text{b}\ce{[B]_\text{eq}}\\ K_\text{c} = \dfrac{k_\text{f}}{k_\text{b}} = \dfrac{\ce{[B]_\text{eq}}}{\ce{[A]_\text{eq}}} $$
Free Energy
Since both processes are spontaneous, $\Delta G$ for both processes, from nonequilibrium state to the equilibriums state, is negative. Free energy is minimum at equilibrium.
