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Chemistry Review Manual 2

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(a) Solid calcium carbonate and calcium oxide at 700°C are in a vessel with 0.1 atm partial<br />

pressure carbon dioxide. At 700°C, K = 0.056 for the decomposition of CaCO 3 (s) into CaO(s)<br />

and CO 2 (g).<br />

(b)<br />

Consider the following equilibrium:<br />

Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 4 2+ (aq) K = 6.7×10 12 at 25°C<br />

Suppose [Cu 2+ ] = [Cu(NH 3 ) 4 2+ ] = 1.0 mol L −1 , while [NH 3 ] = 5.0×10 −4 mol L −1 at 25°C.<br />

(c)<br />

Consider the following equilibrium:<br />

SnO 2 (s) + 2 CO(g) Sn(s) + 2 CO 2 (g)<br />

K = 11 atm at some T<br />

Suppose the partial pressure of CO 2 is 3 atm, while that of CO is 1 atm at the required T.<br />

Approach: Compute Q and compare it with K.<br />

(a) Q = p<br />

CO 2<br />

= 0.1 > 0.056. Therefore, there is net reverse reaction. At this partial<br />

pressure, CO 2 in the gas phase combines with available CaO(s) to form CaCO 3 (s).<br />

(b) Here,<br />

( )<br />

2+<br />

3 4<br />

13<br />

1.6 10<br />

2+<br />

4<br />

4<br />

3<br />

[Cu NH ] 1.0<br />

Q = = = × > K<br />

[Cu ][NH ] 1.0 5.0 10<br />

Again, Q > K and there is net reverse reaction.<br />

−4<br />

( × )<br />

(c) Here,<br />

2 2<br />

pCO<br />

3<br />

2<br />

Q = = = 9<<br />

11<br />

p<br />

2<br />

CO<br />

1<br />

In this case, Q < K and there is net forward reaction.<br />

5.4 Shift in equilibrium – le Châtelier's principle<br />

Non-equilibrium conditions can result, starting with an equilibrium mixture, by adding or removing<br />

a reactant or a product. For example, if we have an equilibrium solution with the reaction<br />

CO 3 2− (aq) + H 2 O(l) HCO 3 − (aq) + OH − (aq)<br />

K = 2.1×10 −4<br />

and we dissolve a small amount of NaOH(s) into the solution, a non-equilibrium initial state<br />

results. Increasing the concentration of hydroxide makes Q larger (a factor in the numerator is<br />

increased). Since Q is now larger than K, there is net reverse reaction – product concentrations<br />

decrease, while the reactant concentration increases. Net reaction continues until Q = K is<br />

restored. When equilibrium is re-established, the hydroxide concentration is smaller than it was<br />

immediately after the added NaOH(s) dissolves, but larger than it was before the NaOH(s) was<br />

added.<br />

We can also add a small amount of concentrated sulfuric acid (so as not to significantly change<br />

the total volume) to the solution. The added strong acid reacts completely with OH − present, and<br />

reduces the hydroxide concentration. As such, it removes a product from the chemical reaction.<br />

This has the effect of making Q smaller. Net forward reaction results until equilibrium is reestablished.<br />

Page 44 of 88

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