Return to Answer

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards.  $$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

The effect of temperature I will add after...

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards. 

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.

If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left.

If $\Delta H>0$, you need to think about heat not as "product", but as "reagent".

For better understanding try to "dive" into Equilibrium thermodynamics.

• $\upsilon'(A)=2\,mol-2x\,mol$

• $\upsilon'(B)=3\,mol-3x\,mol$

• $\upsilon'(C)=1\,mol+x\,mol$

• $\upsilon'(D)=1\,mol+x\,mol$

• $X'(A)=(2-2x)/(7-3x)$

• $X'(B)=(3-3x)/(7-3x)$

• $X'(C)=(1+x)/(7-3x)$

• $X'(D)=(1+x)/(7-3x)$

Now, the partial pressures are (pressure in the jat increased, because quantatiy of molecules increased, but volume stays the same):

• $\upsilon'(A)=2\,mol-2x\,mol$
• $\upsilon'(B)=3\,mol-3x\,mol$
• $\upsilon'(C)=1\,mol+x\,mol$
• $\upsilon'(D)=1\,mol+x\,mol$

Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$

• $X'(A)=(2-2x)/(7-3x)$
• $X'(B)=(3-3x)/(7-3x)$
• $X'(C)=(1+x)/(7-3x)$
• $X'(D)=(1+x)/(7-3x)$

Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same):

The partial pressures are: $$\rm P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$$ $$\rm P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$$ $$\rm P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$$

Now, the partial pressures are: $$\rm P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$$ $$\rm P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$$ $$\rm P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$$ As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.

Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.

• $P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
• $P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
• $P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
• $P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$

Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$ $$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$

Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:

$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$ where

• $\upsilon'(A)=2\,mol-2x\,mol$

• $\upsilon'(B)=3\,mol-3x\,mol$

• $\upsilon'(C)=1\,mol+x\,mol$

• $\upsilon'(D)=1\,mol+x\,mol$

• $X'(A)=(2-2x)/(7-3x)$

• $X'(B)=(3-3x)/(7-3x)$

• $X'(C)=(1+x)/(7-3x)$

• $X'(D)=(1+x)/(7-3x)$

Solving the eqution gives $x=0.325\,mol$, so now:

• $\upsilon'(A)=1.35\,mol$
• $\upsilon'(B)=2.03\,mol$
• $\upsilon'(C)=1.33\,mol$
• $\upsilon'(D)=1.33\,mol$

But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right.

• $P'(A)=0.45\,atm$
• $P'(B)=0.67\,atm$
• $P'(C)=0.44\,atm$
• $P'(D)=0.44\,atm$

The partial pressures are:

• $ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
• $ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
• $P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$

Now, the partial pressures are (pressure in the jat increased, because quantatiy of molecules increased, but volume stays the same):

• $P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
• $P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
• $P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$

As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.