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Calculating K c values

Calculating K c values. CH 3 COOH (aq) + CH 3 CH 2 OH (aq). CH 3 COOCH 2 CH 3 (ag) + H 2 O (l). From the balanced equation 1 mol of acid reacts with 1 mol of alcohol to form 1 mol of ester and 1 mol of water. This means that…. The amount of the two products formed is the same ( x).

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Calculating K c values

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  1. Calculating Kc values CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) From the balanced equation 1 mol of acid reacts with 1 mol of alcohol to form 1 mol of ester and 1 mol of water. This means that….. • The amount of the two products formed is the same (x) • The amount by which the two reactants are reduced equals the amount of each product formed (x)

  2. Calculating Kc values CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) Kc = [CH3COOCH2CH3] [H2O] [CH3COOH ][CH3CH2OH] Kc = (x/V)x (x/V)(a-x)/Vx(b-x)/V Where V = volume in dm3 If the number of moles is the same on both sides of the equilibrium then the volume cancels, so we can use moles rather than concentrations Kc = ___x2___(a-x)x (b-x)

  3. Calculating Kc values CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) Moles CH3COOH at equilibrium = 1.0 – 0.171 = 0.829 Moles CH3CH2OH at equilibrium = 0.18 – 0.171 = 0.009

  4. Calculating Kc values CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) Kc = [CH3COOCH2CH3] [H2O] [CH3COOH ][CH3CH2OH] Kc = 0.171 x 0.171 0.829 x 0.009 = 0.0292 0.829 x 0.009 = 3.92

  5. When 1 mol each of ethanoic acid and ethanol are mixed together at a fixed temperature 0.333mol of acid remain at equilibrium. Calculate Kc CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) Moles CH3CH2OH at equilibrium = 0.333 Moles CH3COOCH2CH3 at equilibrium = 1.0 – 0.333 = 0.667 Moles H2O at equilibrium = 1.0 – 0.333 = 0.667

  6. Calculating Kc values CH3COOH (aq) + CH3CH2OH(aq) CH3COOCH2CH3(ag) + H2O(l) Kc = [CH3COOCH2CH3] [H2O] [CH3COOH ][CH3CH2OH] Kc = 0.667 x 0.667 0.333 x 0.333 = 4.01

  7. 0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate Kc H2(g) + I2g) 2HI(g) From the balanced equation 1mol hydrogen reacts with 1 mol iodine to form 2 moles of hydrogen iodide Moles H2 at equilibrium = 0.206 – 0.258/2 = 0.077 Moles I2 at equilibrium = 0.144 – 0.258/2 = 0.015

  8. 0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate Kc H2(g) + I2g) 2HI(g) Kc = [HI]2 [H2][I2] Kc = (0.258)2 0.077 x 0.015 Kc = 57.63

  9. For reactions that involve a change in the number of moles, the volume must be known in order to calculate the concentrations before Kc can be calculated as the volume terms will not cancel.

  10. At 300K 1.0 mol of N2O4 is 20% dissociated in 2.0dm3 flask. Calculate Kc N2O4(g) 2NO2(g) If 20% is dissociated, then 80% remains at equilibrium Mols of N2O4 at equilibriuum = 80/100 x 1.0 = 0.8 Moles of NO2 at equilibrium = 2 x (1.0 – 0.8) = 0.4

  11. At 300K 1.0 mol of N2O4 is 20% dissociated in 2.0dm3 flask. Calculate Kc N2O4(g) 2NO2(g) Kc = [NO2]2 [N2O4] Kc = (moldm-3)2 moldm-3 = moldm-3 Kc = (0.4/V)2 0.8/V Kc = (0.4/2)2 0.8/2 Kc = 0.1 moldm-3

  12. Finding Kc Experimentally • Known amounts of reagents are used (Initial concentrations known) • The system is closed and left until equilibrium is reached • The concentration of the products is then analysed, often by titration • Kc can then be calculated

  13. Different initial concentrations of alcohol were used in different experiments • Calculate Kc for the following experiments • All the experiments were performed at 373 k

  14. Kc is a constant at constant temperature Altering the concentration of the reactants will shift the equilibrium position but the value of Kc remains constant

  15. Factors that Affect Equilibrium • Changing the concentration of a reactant or product will shift the position of an equilibrium CH3COOCH2CH3(ag) + H2O(l) CH3COOH (aq) + CH3CH2OH(aq) Kc = [CH3COOCH2CH3] [H2O] [CH3COOH ][CH3CH2OH] Kc is the ratio of concentrations so… • Changing the concentration of a reactant or product will not affect the value of Kc

  16. Factors that Affect Equilibrium • Changing concentration has no effect on Kc • Changing pressure has no effect on Kc • Catalysts have no effect on Kc • Changing temperature will change Kc

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