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Chapter 17.  Most reactions do not proceed to completion.  N 2 (g) + H 2 (g)  2NH 3 (g)  2NH 3 (g)  N 2 (g) + H 2 (g)

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Presentation on theme: "Chapter 17.  Most reactions do not proceed to completion.  N 2 (g) + H 2 (g)  2NH 3 (g)  2NH 3 (g)  N 2 (g) + H 2 (g)"— Presentation transcript:

1 Chapter 17

2  Most reactions do not proceed to completion.  N 2 (g) + H 2 (g)  2NH 3 (g)  2NH 3 (g)  N 2 (g) + H 2 (g)

3  Equilibrium is when the concentrations of the reactants and products of a chemical reaction are constant.  Chemical equilibrium happens when the rate of the forward reaction is equal to the rate of the reverse reaction.

4  aA + bB cC + dD  The equilibrium constant expression:  If K eq > 1 the products are favored  If K eq < 1 the reactants are favored

5  Homogeneous:  CO 2 (g)  2CO(g) + O 2 (g)  Heterogeneous:  H 2 O(l)  H 2 O(g)  CaCO 3 (s)  CaO(s) + CO 2 (g)

6  Write the equilibrium constant expression fo the following reactions  N 2 (g) + H 2 (g)  2NH 3 (g)  N 2 O 4 (g)  2NO 2 (g)  2H 2 S(g)  2H 2 (g) + S 2 (g)  CO(g) + 3H 3 (g)  CH 4 (g) + H 2 O(g)  4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g)

7  When equilibrium is reached the concentrations of the reactants and the products will not change.  At any given temperature the value of k eq will always be the same no matter what the equilibrium concentrations are.

8  Calculate the value of k eq for the following reaction and concentrations.  N 2 + 3H 2  2NH 3  [N 2 ] = 0.533 M  [H 2 ] = 1.600 M  [NH 3 ] = 0.933 M  0.399

9  Calculate k eq for the reaction:  N 2 O 4 (g)  2NO 2 (g) If [N 2 O 4 ] = 0.0185 mol/L and [NO 2 ] = 0.0627 mol/L  Calculate k eq for the reaction  CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) if [CO] = 0.0613 mol/L, [H 2 ] = 0.1839 mol/L, [CH 4 ] = 0.0387 mol/L and [H 2 O] = 0.0387 mol/L  For the reaction COCl 2 (g)  CO(g) + Cl 2 (g)

10  The by-products of an industrial process are CO and H 2. These two gasses can combine to produce CH 4 and H 2 O in equilibrium.  Use the following data to find k eq for this reaction.  [CO] = 0.300M, [H 2 ] = 0.100M, [CH 4 ] = 0.059 M, [H 2 O] = 0.020 M  The chemists in charge of this industrial process would like to make use of the methane (CH 4 ) that is being produced. In order for them to have a usable amount of methane the concentration must be 0.100 M or higher. How can they achieve this?

11  If a stress is applied to a system at equilibrium, the system will shift in the direction that relieves that stress.  Suppose additional CO is injected into the reaction vessel of the industrial process we discussed. How would this affect the equilibrium of the system?

12  Along with adding reactants or removing products, what other stresses can we apply to a system?  Temperature  Pressure

13  The reaction:  CO(g) + 3 H 2 (g)  CH 4 (g) + H 2 O(g)  Has a K eq of 3.933 at 1200 K.  If the [CO] = 0.850 M, [H 2 ] = 1.33 M, and [H 2 O]= 0.286 M, what is the concentration of CH 4 ?

14  Some ionic compounds dissolve completely in water.  Ex: NaCl(aq)  Na + (aq) + Cl - (aq)  Some ionic compounds do not dissolve completely in water.  Ex: BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq)  This process is happening in equilibrium.  So there is an equilibrium constant associated with it.

15  Calculate the solubility of AgI(s).  k sp = 8.5 x 10 -17

16  Examples  Calculate the solubility in mol/L of CuCO 3 if it’s k sp is 2.5 x 10 -10  1.6 x 10 -5 mol/L  Calculate the solubility PbCrO 4 if it’s k sp is 2.3x10 -13.  4.8 x 10 -7 mol/L  Calculate the solubility of CaF 2 if it’s k sp is 3.5x10 -11.  4.18x10 -6 mol/L

17  Sometimes the concentrations of the ions are not the same as the solubility.  Mg(OH) 2 (s)  Mg 2+ (aq) + 2OH - (aq)  k sp = 5.6 x 10 -12


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