5 How the reaction equation affects K It is important to remember that an equilibrium quotient or constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of Q or K will change. The rules are very simple: • Writing the equation in reverse will invert the equilibrium expression; • Multiplying the coefficients by a common factor will raise Q or K to the corresponding power. Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements: 2 H2 + O2→ 2 H2O 10 H2 + 5 O2 → 10 H2O H2 + ½ O2 → H2O H2O → H2 + ½ O2 Equilibrium constant for a sequence of reactions Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to this rule: The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps. Problem Example 2 Given the following equilibrium constants: CaCO3(s) → Ca2+(aq) + CO32–(aq) K1 = 10–6.3 HCO3–(aq) → H+(aq) + CO32–(aq) K2 = 10–10.3 Calculate the value of K for the reaction CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3–(aq) Solution: The net reaction is the sum of reaction 1 and the reverse of reaction 2: CaCO3(s) → Ca2+(aq) + CO32–(aq) K1 = 10–6.3 H+(aq) + CO32–(aq) → HCO3–(aq) K–2 = 10–(–10.3) CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3–(aq) K = K1/K2 = 10(-8.4+10.3) = 10+1.9 Comment: This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid rain on buildings and statues. This an example of a reaction that has practically no tendency to take place by itself (small K1) being driven by a second reaction having a large equilibrium constant (K–2). From the standpoint of the LeChâtelier principle, the first reaction is pulled to the right by the removal of carbonate by hydrogen ion. Coupled reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked. Problem Example 3 The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant Kp = 4.5¥1015 at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find Kp for the homogeneous gas-phase reaction at the same temperature. Solution: The net reaction we seek is the sum of the heterogeneous synthesis of HBr and the reverse of the vaporization of liquid bromine: H2(g) + Br2(l) → 2 HBr(g) Kp = 4.5×1015 Br2(g) → Br2(l) Kp = (0.28)–1 H2(g) + Br2(g) → 2 HBr(g) Kp = 1.6×1019
Posted on: Wed, 19 Nov 2014 19:02:21 +0000
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