5.1.2 (f,g) How Rate Constants change with changing conditions.
Syllabus
f(i) the qualitative effect on equilibrium constants of changing temperature for exothermic and endothermic reactions
(ii) the constancy of equilibrium constants with changes in concentration, pressure or in the presence of a catalyst
g)explanation of how an equilibrium constant controls the position of equilibrium on changing concentration, pressure and temperature
What does this mean?
Pressure & Kc
You should already know that Kc is unaffected by changes in:
- Pressure
- Addition of catalyst
It seems counter-intuitive that pressure would not change the position of an equilibrium involving unequal numbers of moles of reactant and product gases. But this is because we tend to confuse moles with concentration.
eg. 2SO2(g) + O2(g) ⇌ 2SO3(g)
If we increase the pressure then the equilibrium will certainly move to the right because there are fewer gas moles on the product side.
But Kc = [SO3]2 / [SO2]2[O2]
But the [] symbols are not moles.
They are concentrations.
In other words, moles/volume.
Kc = (nSO3/V)2 /(nSO2/V)2(nO2/V)
And we have increased the pressure.
So the volume will decrease.
And the conventional explanation that the examiner will accept is that the effect on the concentrations cancel out.
Pressure & Kp
The equivalent for Kp is:
Kp = p2SO3 / p2SO2 x pO2
But partial pressure is given by pA = XA x pTotal (XA = Mole Fraction of A)
So,
Kp = (XSO3 x p)2/ (XSO2 x p)2 x (XO2 x p)
Kp = XSO32/ XSO22 x (XO2 x p)
So the mole fraction of SO3 (XSO3) will increase while the mole fractions of SO2 and O2 decrease.
But the pressure increased so the effect again cancels out - which is all examiners ever want from you.
Effect of Catalysts.
Catalysts don't change the position of equilibria.
They simply get us to the same equilibrium faster.
Hence, catalysts don't affect Kc or Kp.
The catalyst decreases the Activation Energy for both the forward and backward reactions.
Once again, the effects cancel each other out.
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